Sequences and Series: An Introduction for Beginners

Abstract

This mathematics book presents the main definitions and properties of sequences, series of real numbers, sequences, and series of functions. We combine a rigorous exposition with a set of intuitive examples to achieve an easily readable final text that is approachable by intermediate and more advanced calculus students from the secondary to the university level. This book is divided into three sections: an introductory one where we present the basic mathematical concepts and notation we will follow; a section devoted to sequences of real numbers; and a section dedicated to series of real numbers, sequences, and functions. These two last sections end with exercises with solutions for the readers to test their learning and knowledge. Our objective was to provide students and curious minds on the subject of the book with a tool for autonomous learning.

1. Preliminaries and notation

We decided to begin this book with a brief exposition of the main basic mathematical concepts we will use and assume the reader is familiar with. Also, the notation used in this book will be presented in the following paragraphs. For further reading on the topics addressed in this book, we recommend the detailed expositions presented in Refs. [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]. Refs. [1, 2, 3, 4, 5, 6, 7] consist of theoretical fundamental texts while Refs. [8, 9, 10] present more practical approaches.

1.1 Basic set notation

In this book, we will deal with real numbers. Let us start by recalling the basic set notation.

We denote by $\mathbb{N}$ the set of all natural numbers excluding 0, that is,

$$\mathbb{N}=\left\{1,2,3,\dots \right\},$$

and the set of integers is denoted by $\mathbb{Z}$,

$$\mathbb{Z}=\left\{\dots ,-3,-2,-1,0,1,2,3,\dots \right\}.$$

Next, the set of rational numbers is denoted by $\mathbb{Q}$:

$$\mathbb{Q}=\left\{\frac{a}{b}:a,b\in \mathbb{Z}\wedge b\ne 0\right\}.$$

Finally, we denote by $\mathbb{R}$ the set of real numbers, that is,

$$\mathbb{R}=\mathbb{Q}\cup \left\{x:x\text{is irrational}\right\},$$

where it is sufficient to say that being irrational is not being able to be written as the division of two integers.

Some additional symbols may be used to represent special subsets of the main sets presented above like, for example,

$$\begin{array}{lll}{\mathbb{N}}_0& =& \mathbb{N}\cup \left\{0\right\};\\ {\mathbb{N}}_p& =& \left\{n\in \mathbb{Z}:n\ge p\right\}\\ {\mathbb{Z}}^{-}& =& \left\{x\in \mathbb{Z}:x<0\right\};\\ {\mathbb{R}}_0^{+}& =& \left\{x\in \mathbb{R}:x\ge 0\right\}.\end{array}$$

1.2 Upper and lower bounds of sets

Some sets of real numbers are limited as their elements will not be bigger or smaller than some values.

Definition 1.1 (Bounded sets). Let X be a non-empty subset of $\mathbb{R}$.

1. X is said to be bounded from above if there is a real number M such that $x\le M,\forall x\in X.$ In these conditions, M is called an upper bound of X.

2. X is said to be bounded from below if there is a real number m such that $x\ge m,\forall x\in X.$ In these conditions, m is called a lower bound of X.

3. X is said to be bounded if it is bounded from above and below and unbounded if it is not bounded.

Definition 1.2 (Maximum, minimum, supremum, and infimum of sets). Let X be a non-empty subset of $\mathbb{R}$. A real number is

1. the maximum of X if it is an element of X and if it is an upper bound of X.

2. the minimum of X if it is an element of X and if it is a lower bound of X.

3. the supremum of X if it is the smaller of all upper bounds of X.

4. the infimum of X if it is the bigger of all lower bounds of X.

Note that, for a given set, the existence of any of the numbers presented in Definition 1.2 is not guaranteed. Also, observe that, when the maximum and minimum of X do exist, they coincide with the supremum and infimum of X, respectively.

Example 1.1. Let $X=]0,1]$. For this set, the following statements are true.

• This set is bounded from above, as $[1,+\infty [$ is the set of all upper bounds of X.

• The set X is bounded from below, as $]-\infty ,0]$ is the set of all lower bounds of X.

• Hence, X is a bounded set.

• The maximum of X is 1.

• X does not have a minimum.

• The supremum of X is 1.

• The infimum of X is 0.

The concept of neighborhood of a number will also be used ahead. We define it in a simple way as follows.

Definition 1.3 (Neighborhood of a number). Let a be a real number. A neighborhood of a is a set that contains an interval $]a-\epsilon ,a+\epsilon [$, with $\epsilon >0$.

1.3 Relations and functions

Given two sets X and Y, the Cartesian product $X\times Y$ is defined as

$$X\times Y=\left\{\left(x,y\right):x\in X\wedge y\in Y\right\}$$

and its elements are called ordered pairs.

A binary relation or correspondence, R, between X and Y is simply a subset of $X\times Y$. That subset is often called the graph of R. If $\left(x,y\right)\in R$, it means that x is related to y through R, we say that x is R-related to y, and we write xRy. The order is important, as xRy does not necessarily imply that yRx.

The sets X and Y are usually called the departure set and the destination set of R, respectively. The domain of R is the set of elements of X that are R-related to at least one element of Y, and the image of R is the set of elements of Y for which there is at least one element of x such that xRy.

There are different types of binary relations.

Definition 1.4 (Types of binary relations). Let R be a binary relation between the sets X and Y.

1. If, for all $x\in X$ and $y_1,y_2\in Y$ we have ${\mathit{xRy}}_1\wedge {\mathit{xRy}}_2\Rightarrow y_1=y_2$, then R is said to be a functional relation.

2. If, for all $x\in X$, there exists at least one $y\in Y$ such that xRy, then R is said to be left-total.

3. If R is functional and left-total, it is called a function.

4. If, for all $x_1,x_2\in X$ and $y\in Y$ we have $x_1\mathit{Ry}\wedge x_2\mathit{Ry}\Rightarrow x_1=x_2$, then R is said to be injective.

5. If, for all $y\in Y$ there exists at least one $x\in X$ such that xRy, then R is said to be surjective or right-total.

6. A function that is injective and surjective is said to be bijective or a bijection.

Example 1.2. Let $X=\left\{1,,,2,,,3\right\},Y=\left\{a,b,c\right\}$_, and $Z=\left\{\alpha ,\beta \right\}$. Then:

• $R_1=\left\{\left(1,a\right),\left(2,b\right)\right\}$ is a functional and injective relation between X and Y that is not a function;

• $R_2=\left\{\left(1,\alpha \right),\left(2,\beta \right)\right\}$ is a functional, injective, and surjective relation between X and Z that is not a function because it is not left-total;

• $R_3=\left\{\left(\alpha ,a\right),\left(\alpha ,b\right),\left(\beta ,c\right)\right\}$ is a left-total, surjective, but not injective, relation between Z and Y that is not a function because it is not functional;

• $R_4=\left\{\left(1,a\right),\left(2,a\right),\left(3,a\right)\right\}$ is function between X and Y that is not injective or surjective;

• $R_5=\left\{\left(\alpha ,a\right),\left(\beta ,b\right)\right\}$ is an injective, but not surjective, function between Z and Y;

• $R_6=\left\{\left(1,\alpha \right),\left(2,\alpha \right),\left(3,\beta \right)\right\}$ is a surjective, but not injective, function between X and Z;

• $R_7=\left\{\left(1,c\right),\left(2,a\right),\left(3,b\right)\right\}$ is a bijective function between X and Y.

Usually, functions are denoted by the letter f and, if x and y are f-related, instead of writing xfy, we write $y=f\left(x\right)$.

1.4 Limits and derivatives of functions

In this section, we make a brief summary of two concepts that might be of interest for the reader to understand a few examples presented in this text.

The concept of limit of a function can actually be formally defined with the help of the limit of sequences, which will be surveyed in detail in Section 2.4 of Section 2. The idea behind both of them is, however, the same. Given a function f, we say that the limit of f, when x approximates to the value a, is b, and we write

$$\underset{x\to a}{lim}f\left(x\right)=b,$$

if the images of the elements of the domain, which are converging to a, are converging to b. Note that a and b can be $\infty$. In the special case of sequences, that we will study in the next section, the limits are always calculated when $a=+\infty$ because of the nature of this special kind of functions.

The derivative of a function, f, of one variable, x, is essentially an instantaneous rate of change of the images of f with the values of x. It is denoted by $f^{\prime }$, Lagrange’s notation, or by $\frac{\mathit{df}}{\mathit{dx}}$, Leibniz’s notation, and can be defined by the limit (if the limit exists):

$$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}.$$E1

If the limit in (1) is a real number, then f is said to be differentiable at x.

From (1), one can easily deduce formulae for the derivative of many functions. Two basic rules that can be deduced are very useful while differentiating are the product and the quotient rules: given two functions f and g, both differentiable, then

$$\begin{array}{lll}{\left(\mathit{fg}\right)}^{\prime }\left(x\right)& =& f^{\prime }\left(x\right)g\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)\\ & \text{and}& \\ {\left(\frac{f}{g}\right)}^{\prime }\left(x\right)& =& \frac{f^{\prime }\left(x\right)g\left(x\right)-f\left(x\right)g^{\prime }\left(x\right)}{g^2\left(x\right)},\end{array}$$

provided, in the latter case, that $g\left(x\right)\ne 0$.

To differentiate other more complex analytic expressions of functions one makes use of the so-called chain rule. Indeed, if we have a composite function $f\circ g$, then we have:

$${\left(f\circ g\right)}^{\prime }\left(x\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right),$$E2

provided that g is differentiable at x and f is differentiable at g(x).

For the reader to be able to follow some of the examples presented ahead, we will provide the most common derivatives in Table 1. All of the formulae presented can be obtained by making use of (1) and/or (2). For more insight on this derivative operator, including its operation properties, please consult any calculus book; as for the scope of this book, what was enlightened is enough, and we do not pretend to go deeper on this subject.

f(x)	$f^{\prime }\left(x\right)$	f(x)	$f^{\prime }\left(x\right)$
a	0	u	$u^{\prime }$
ax	a	au	$a{u}^{\prime }$
${\mathit{ax}}^n$	${\mathit{nax}}^{n-1}$	${\mathit{au}}^n$	${\mathit{anu}}^{n-1}{u}^{\prime }$
$\sqrt{x}$	$\frac{1}{2\sqrt{x}}$	$\sqrt{u}$	$\frac{u^{\prime }}{2\sqrt{u}}$
$\sqrt[n]{x}$	$\frac{1}{n\sqrt[n]{x^{n-1}}}$	$\sqrt[n]{u}$	$\frac{u^{\prime }}{n\sqrt[n]{u^{n-1}}}$
ln(x)	$\frac{1}{x}$	ln(u)	$\frac{u^{\prime }}{u}$
${log}_b\left(x\right)$	$\frac{1}{xlnb}$	${log}_b\left(u\right)$	$\frac{u^{\prime }}{ulnb}$
$e^x$	$e^x$	$e^u$	$u^{\prime }{e}^u$
$b^x$	$b^{x}ln\left(b\right)$	$b^u$	$u^{\prime }{b}^{u}ln\left(b\right)$
sin(x)	cos(x)	sin(u)	$u^{\prime }cos\left(u\right)$
cos(x)	− sin(x)	cos(u)	$-u^{\prime }sin\left(u\right)$
tan(x)	${sec}^2\left(x\right)$	tan(u)	$u^{\prime }{sec}^2\left(u\right)$
arcsin(x)	$\frac{1}{\sqrt{1-x^2}}$	arcsin(u)	$\frac{u^{\prime }}{\sqrt{1-u^2}}$
arccos(x)	$-\frac{1}{\sqrt{1-x^2}}$	arccos(u)	$-\frac{u^{\prime }}{\sqrt{1-u^2}}$
arctan(x)	$\frac{1}{x^2+1}$	arctan(u)	$\frac{u^{\prime }}{u^2+1}$
sinh(x)	cosh(x)	sinh(u)	$u^{\prime }cosh\left(u\right)$
cosh(x)	sinh(x)	cosh(u)	$u^{\prime }sinh\left(u\right)$
tanh(x)	${sech}^2\left(x\right)$	tanh(u)	$u^{\prime }{sech}^2\left(u\right)$

Table 1.

Most common derivatives $(a\in \mathbb{R},b\in {\mathbb{R}}^{+}\backslash \left\{1\right\}$ and u is a function with respect to x).

Finally, we will make use, at some point, of a technique, called l’Hôpital’s Rule, for computing limits of functions that are the division of two other functions. The technique has two cases of application that are presented next, but both work basically in the same way; that is, you just have to consider the derivatives of the numerator function and the denominator function and recalculate the limit. Usually, differentiating a function makes it simpler. This rule made it is first apparition in a book of the French mathematician Guillaume de l’Hôpital.

Theorem 1.1 (l’Hôpital’s Rule for 0/0). Let f and g be two differentiable functions on an open interval I containing $a\in \mathbb{R}$ and g is such that $g^{\prime }\left(x\right)\ne 0$ in $I\backslash \left\{a\right\}.$ If ${lim}_{x\to a}f\left(x\right)={lim}_{x\to a}g\left(x\right)=0$, then

$$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)},$$

so long as the limit is finite or $\pm \infty$.

Theorem 1.2 (l’Hôpital’s Rule for $\infty /\infty$). Let f and g be two differentiable functions on an open interval I containing $a\in \mathbb{R}$ and g is such that $g^{\prime }\left(x\right)\ne 0$ in $I\backslash \left\{a\right\}$. If ${lim}_{x\to a}f\left(x\right)=\pm \infty$ and ${lim}_{x\to a}g\left(x\right)=\pm \infty$, then

$$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)},$$

so long as the limit is finite or $\pm \infty$.

Note that, in both cases, similar results hold for $a=\pm \infty$.

In the following, we present some practical examples of the effectiveness of l’Hôpital’s Rule on calculating limits.

Example 1.3. Let’s use l’Hôpital’s Rule to compute

$$\underset{x\to 0}{lim}\frac{e^x+2}{3x}.$$

Indeed, we have

$$\begin{array}{lll}\underset{x\to 0}{lim}\frac{2-2e^x}{3x}& \stackrel{\left(\frac{0}{0}\right)}{=}& \underset{x\to 0}{lim}\frac{\frac{d}{\mathit{dx}}\left(2-2e^x\right)}{\frac{d}{\mathit{dx}}\left(3x\right)}\\ & =& \underset{x\to 0}{lim}\frac{-2e^x}{3}\\ & =& \frac{-2e^0}{3}\\ & =& -\frac{2}{3}.\end{array}$$

Example 1.4. Now we will compute

$$\underset{x\to 0}{lim}\frac{sinx}{5x}.$$

using l’Hôpital’s Rule. We have

$$\begin{array}{lll}\underset{x\to 0}{lim}\frac{sinx}{5x}& \stackrel{\left(\frac{0}{0}\right)}{=}& \underset{x\to 0}{lim}\frac{\frac{d}{\mathit{dx}}\left(sinx\right)}{\frac{d}{\mathit{dx}}\left(5x\right)}\\ & =& \underset{x\to 0}{lim}\frac{cosx}{5}\\ & =& \frac{1}{5}.\end{array}$$

Example 1.5. In this example, we will apply l’Hôpital’s Rule to calculate

$$\underset{x\to +\infty }{lim}\frac{e^{x-1}}{2x^3}.$$

We have

$$\begin{array}{lll}\underset{x\to +\infty }{lim}\frac{e^{x-1}}{2x^3}& \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& \underset{x\to +\infty }{lim}\frac{\frac{d}{\mathit{dx}}\left(e^{x-1}\right)}{\frac{d}{\mathit{dx}}\left(2x^3\right)}\\ & =& \underset{x\to +\infty }{lim}\frac{e^{x-1}}{6x^2}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& \underset{x\to +\infty }{lim}\frac{\frac{d}{\mathit{dx}}\left(e^{x-1}\right)}{\frac{d}{\mathit{dx}}\left(6x^2\right)}\\ & =& \underset{x\to +\infty }{lim}\frac{e^{x-1}}{12x}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& \underset{x\to +\infty }{lim}\frac{\frac{d}{\mathit{dx}}\left(e^{x-1}\right)}{\frac{d}{\mathit{dx}}\left(12x\right)}\\ & =& \underset{x\to +\infty }{lim}\frac{e^{x-1}}{12}\\ & =& \frac{+\infty }{12}\\ & =& +\infty .\end{array}$$

1.5 Mathematical induction

Since sequences are essentially, as we will shortly see, functions with domain $\mathbb{N}$, it is important to recall the principle of mathematical induction, an important technique for proving properties among the naturals.

Proposition 1.1 (Principle of Mathematical Induction). Let P(n) be a statement involving the natural number n. The proposition

$$\forall n\in \mathbb{N},P\left(n\right)$$

is true if:

1. the statement P(n) is true for $n=1$, that is, P(1) is true;

2. if the statement P(n) is true for $n=m\in \mathbb{N}$, then it is also true for $n=m+1$, that is,

$$\forall m\in \mathbb{N},P\left(m\right)\Rightarrow P\left(m+1\right).$$

The principle of mathematical induction comes from the axiomatic definition of $\mathbb{N}$ which is defined as the smallest subset, S, of $\mathbb{R}$ such that, if $x\in S$, then $x+1\in S$. If the statement has this inductive property, that is, if it is valid for a number m, then it is also valid for it is successor, $m+1,$ then verifying that the statement is valid for 1 originates a chain reaction that proves the statement for all-natural numbers.

Note that Proposition 1.1 can be generalized for sets ${\mathbb{N}}_p$, with $p\in \mathbb{N}$. In this case, the first step is to verify if the statement is valid for $n=p$ and not $n=1$. The rest is analogous.

Example 1.6. Using mathematical induction, let’s prove that, for each $n\in \mathbb{N}$, we have

$$1^2+2^2+3^2+\cdots +n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}.$$

For $n=1$, we have only one element in the sum, therefore,

$$\begin{array}{l}{1}^2=\frac{1\times \left(1+1\right)\times \left(2\times 1+1\right)}{6}\\ \iff 1=\frac{1\times 2\times 3}{6}\\ \iff 1=1,\end{array}$$

and the formula is valid.

Now, suppose that, for a certain $m\in \mathbb{N}$, we have

$$1^2+2^2+\cdots +m^2=\frac{m\left(m+1\right)\left(2m+1\right)}{6}.$$

Under this assumption (usually called the hypothesis of induction), let’s prove that the formula is valid for $m+1$ (usually called the thesis of induction), that is, that is valid the equality

$$\begin{array}{lll}{1}^2+2^2+\cdots +{\left(m+1\right)}^2& =& \frac{\left(m+1\right)\left(m+2\right)\left(2m+3\right)}{6}\\ & =& \frac{2m^3+9m^2+13m+6}{6}.\end{array}$$

Indeed, using the hypothesis of induction, we have

$$\begin{array}{lll}{1}^2+2^2+\cdots +m^2+{\left(m+1\right)}^2& =& \frac{m\left(m+1\right)\left(2m+1\right)}{6}+{\left(m+1\right)}^2\\ & =& \frac{m\left(m+1\right)\left(2m+1\right)+6{\left(m+1\right)}^2}{6}\\ & =& \frac{2m^3+3m^2+m+6m^2+12m+6}{6}\\ & =& \frac{2m^3+9m^2+13m+6}{6}\\ & =& \frac{\left(m+1\right)\left(m+2\right)\left(2m+3\right)}{6}.\end{array}$$

This means that, if the formula is valid for $m\in \mathbb{N}$, then it is also valid for $m+1$. Since we already proved that the formula is valid for $n=1$, we can conclude, by mathematical induction, that the formula is valid for all $n\in \mathbb{N}$.

Example 1.7. In this example, we will prove, using induction, that for each natural n, we have:

$$\left(n+3\right)!\ge 2^{n+3}.$$E3

Indeed, for $n=1$, we have

$$\left(1+3\right)!=24\ge 16=2^{1+3},$$

which means that (3) is valid.

Let’s suppose that, for a certain natural m, we have

$$\left(m+3\right)!\ge 2^{m+3}.$$

Let’s prove that

$$\left(m+1+3\right)!\ge 2^{m+1+3}\iff \left(m+4\right)!\ge 2^{m+4}.$$

We have:

$$\begin{array}{lll}\left(m+4\right)!& =& \left(m+4\right)\left(m+3\right)!\\ & \ge & \left(m+4\right)2^{m+3}\\ & >& 2\times 2^{m+3}\\ & =& 2^{m+4},\end{array}$$

using the induction hypothesis in the second step and, in the third step, the fact that $m+3>2.$Hence,

$$\left(m+4\right)!\ge 2^{m+4}$$

and we can conclude, by induction, that (3) is valid for all n in $\mathbb{N}$.

2. Sequences

In this section, we present a general survey on the topic of sequences. A sequence is just a function of arbitrary real numbers but whose domain is the set of natural numbers, $\mathbb{N}$. The origin of such functions can be traced back farther until at least ancient Greece, particularly to the grounds of the mythical Pythagoras’ school, sixth century BC, and the practice of its followers.

This special kind of functions has many practical applications, since many real-life situations are mainly a discrete succession of outputs, which can be modeled by a sequence of real numbers. For example, consider the sequence of perfect squares, also known simply as the sequence of squares: 1, 4, 9, 16, 25, 36, and so on. This sequence is the function of domain $\mathbb{N}$ that takes each natural number and maps it into its square.

Additionally, sequences are able to open up a window toward some rather complex mathematical concepts, such as the different types of infinity, as well as the concepts of convergence and limit. In particular, sequences allow us to establish, in a rather easy and elegant manner, the concept of limit, which is a structural stone in the field of mathematical analysis.

This section is organized in the following way. First, in Section 2.1, we present some general properties of sequences such as monotony and bounds. In Sections 2.2 and 2.3, we present and characterize a couple of special kinds of sequences, which are the arithmetic and the geometric progressions, respectively. Then, in Section 2.4, we address the fundamental problem of sequence convergence, presenting the most relevant results in the subject and examples, and finally, in Section 2.5, we present useful techniques for computing the limits of sequences. We finish the section with some suggested exercises in Section 2.6.

2.1 General properties of sequences

We will start this section with a formal definition of sequence.

Definition 2.1 (Sequence). Let A be a non-empty set. A sequence, u, in A is any function that can be defined from $\mathbb{N}$ into A, that is,

$$\begin{array}{llll}u:& \mathbb{N}& \to & A\\ & n& \mapsto & u\left(n\right)\end{array}.$$

The function values of the sequence u, that is, u(1), u(2), u(3), and so on, are called the terms of the sequence u, with u(1) being the first term, u(2) the second term, u(3) the third term, etc. The set of terms of a sequence u is precisely the image of the map u. The object of a term is also referred to as the order of the term. For instance, the fourth term of the sequence, u(4), is the term of order 4, and the nth term, u(n), is the term of order n.

Along this text, we will consider only real sequences, that is, sequences defined in $\mathbb{R}$. We will also adopt the following standard notation for sequences. Instead of writing u(1), u(2), … , u(n),… , we will denote the terms of the sequence as $u_1,u_2,\dots ,u_n,\dots$, and the sequence u, itself, will be represented as ${\left(u_n\right)}_{n\in \mathbb{N}}$, or simply by $\left(u_n\right)$.

In frequent cases, the term of order n of a sequence can be obtained by a formula involving n. Let us take another look at the example of the square numbers presented at the beginning of this section.

Example 2.1. Consider the sequence, $\left(u_n\right),$ of squares number (see Figure 1).

Figure 1.

Figure of squares.

The term of order n of this sequence is given by $u_n=n^2$.

In cases similar to Example 2.1, where the term of order n is given by a certain formula involving n, $u_n$ assumes a greater level of importance within the sequence $\left(u_n\right)$ because it basically yields an algebraic expression that is capable of generating all the terms of $\left(u_n\right)$, given the value of n. In view of this, the term of order n of $\left(u_n\right)$ is often referred to as the general term of the sequence.

The terms of a sequence $\left(u_n\right)$ can also be plotted out in a Cartesian coordinate system. The graph of the sequence $\left(u_n\right)$ is the set

$$\left\{\left(1,u_1\right),(2,u_2),(3,u_3),\dots ,(n,u_n),\dots \right\}.$$

Example 2.2. Consider the sequence $\left(v_n\right)$ defined by its general term

$$v_n=\frac{3n+2}{n}.$$

The graph of $\left(v_n\right)$ is the set

$$\left\{\left(1,5\right),(2,4),\left(3,\frac{11}{3}\right),\left(4,\frac{7}{2}\right),\left(5,\frac{17}{5}\right),\dots \right\},$$

which can be graphed as in Figure 2:

Figure 2.

Graphically representation of a succession 2.

One can also define a sequence by recurrence, that is, after defining one, or more, of the terms of the sequence, all the other terms are obtained by solving a so-called recurrence equation involving terms already determined before. In general, each term is obtained by making use of the term or terms that precede it. A famous example of a sequence defined in such a manner is the sequence of the Fibonacci numbers.

Example 2.3. The Fibonacci numbers are the following: 1, 1, 2, 3, 5, 8, 13, 21, and so on. After the first two numbers, each new number is obtained by adding the previous two Fibonacci numbers. This sequence of numbers, ${\left(F_n\right)}_{n\in \mathbb{N}}$, can be defined as follows:

$$\left\{\begin{array}{l}{F}_1=1\\ F_2=1\\ F_n=F_{n-1}+F_{n-2},\forall n\in \mathbb{N}\backslash \left\{1,2\right\}\end{array}\right.$$

All the basic algebraic operations for real numbers are still valid for real sequences. Indeed, if ${\left(a_n\right)}_{n\in \mathbb{N}}$ and ${\left(b_n\right)}_{n\in \mathbb{N}}$ are two real sequences, then one can define the sequence of the sum, ${\left(a+b\right)}_{n\in \mathbb{N}}$, the sequence of the difference, ${\left(a-b\right)}_{n\in \mathbb{N}}$, the sequence of the product, ${\left(\mathit{ab}\right)}_{n\in \mathbb{N}}$, and the sequence of the quotient, ${\left(\frac{a}{b}\right)}_{n\in \mathbb{N}}$, providing, in the latter case, that $b_n\ne 0,\forall n\in \mathbb{N}$, whose general terms are defined as ${\left(a+b\right)}_n=a_n+b_n,{\left(a-b\right)}_n=a_n-b_n,{\left(\mathit{ab}\right)}_n=a_n{b}_n$ and ${\left(\frac{a}{b}\right)}_n=\frac{a_n}{b_n},$ respectively.

We proceed the exposition with some general properties of some sequences.

Definition 2.2 (Monotonic sequence). A sequence $\left(u_n\right)$ is:

1. increasing if, $\forall n\in \mathbb{N},u_n\le u_{n+1};$

2. decreasing if, $\forall n\in \mathbb{N},u_n\ge u_{n+1};$

3. strictly increasing if, $\forall n\in \mathbb{N},u_n<u_{n+1};$

4. strictly decreasing if, $\forall n\in \mathbb{N},u_n>u_{n+1};$

5. monotonic if it is increasing or decreasing;

6. strictly monotonic if it is strictly increasing or strictly decreasing.

We follow up Definition 2.2 with some examples.

Example 2.4. Consider the sequence $\left(w_n\right)$ with general term given by

$$w_n=2-3n.$$

Since

$$\begin{array}{lll}{w}_{n+1}-w_n& =& 2-3\left(n+1\right)-\left(2-3n\right)\\ & =& 2-3n-3-2+3n\\ & =& -3<0,\end{array}$$

we conclude that $w_n>w_{n+1}$ holds for each $n\in \mathbb{N}$, thus $\left(w_n\right)$ is strictly decreasing.

Example 2.5. Let $\left(u_n\right)$ be the sequence whose general term is

$$u_n=\frac{2}{2-n}.$$

In this case, we have

$$\begin{array}{lll}{u}_{n+1}-u_n& =& \frac{2}{2-\left(n+1\right)}-\frac{2}{2-n}\\ & =& \frac{2}{1-n}-\frac{2}{2-n}\\ & =& \frac{2\left(2-n\right)-2\left(1-n\right)}{\left(1-n\right)\left(2-n\right)}\\ & =& \frac{2}{\left(1-n\right)\left(2-n\right)}>0,\end{array}$$

since the formula in the denominator always yields positive numbers, regardless of the value of n. Therefore, we can conclude that $u_n<u_{n+1}$ holds for each $n\in \mathbb{N}$, thus $\left(u_n\right)$ is strictly increasing.

Example 2.6. In this example, consider $\left(v_n\right)$ to be the sequence defined by

$$v_n=4n+2{\left(-1\right)}^n.$$

For this sequence, we have

$$\begin{array}{lll}{v}_{n+1}-v_n& =& 4\left(n+1\right)+2{\left(-1\right)}^{n+1}-\left(4n+2{\left(-1\right)}^n\right)\\ & =& 4+2\left({\left(-1\right)}^{n+1}-{\left(-1\right)}^n\right).\end{array}$$

Observe that this expression yields the values 0 and 8, successively, when n is even or odd, respectively. This means that $v_n=v_{n+1}$ when n is even and $v_n<v_{n+1}$ when n is odd. Accordingly, $\left(v_n\right)$ is an increasing, but not strictly increasing, sequence.

Example 2.7. To finish this set of examples, consider the sequence $\left(w_n\right)$, whose general term is given by

$$w_n=n^2-6n.$$

In this case, we have

$$\begin{array}{lll}{w}_{n+1}-w_n& =& {\left(n+1\right)}^2-6\left(n+1\right)-\left(n^2-6n\right)\\ & =& n^2+2n+1-6n-6-n^2+6n\\ & =& 2n-5.\end{array}$$

This expression yields negative values for $n\in \left\{1,2\right\}$ and positive values in the remaining cases. This means that $w_n>w_{n+1}$ when $n=1,2$ and $w_n<w_{n+1}$ when $n>2$. Accordingly, $\left(w_n\right)$ is a non-monotonic sequence.

In the following, we introduce the concept of bounded sequences.

Definition 2.3 (Bounded sequence). A sequence $\left(u_n\right)$ is:

1. bounded from above if there is a $M\in \mathbb{R}$, such that $u_n\le M,\forall n\in \mathbb{N};$

2. bounded from below if there is a $m\in \mathbb{R}$, such that $u_n\ge m,\forall n\in \mathbb{N};$

3. bounded if it is bounded from above and below.

From Definition 2.3, we conclude that a bounded sequence, $\left(u_n\right)$, is one in which all terms are bounded by certain real numbers m and M, that is, $m\le u_n\le M$. It is straightforward to see, though, that this assertion can be slightly modified. In fact, considering $k=max\left\{|m|,|M|\right\}$, we have that $\mid u_n\mid \le k,\forall n\in \mathbb{N}$.

There is a couple of immediate consequences relating to Definitions 2.2 and 2.3. An increasing sequence is clearly bounded from below by its first term, and a decreasing sequence is clearly bounded from above also by its first term.

One might be tempted to conclude that a strictly increasing sequence cannot be bounded from above, but such a conclusion is false. The same applies to strictly decreasing sequences bounded from below. The classical examples are the sequences of general terms $u_n=-\frac{1}{n}$ and $v_n=\frac{1}{n}$. The sequence $\left(u_n\right)$ is strictly increasing and $\left(v_n\right)$ is strictly decreasing, yet we have $-1\le u_n\le 0$ and $0\le v_n\le 1$, thus both sequences are bounded.

Example 2.8. Consider the sequence $\left(w_n\right)$, whose general term is

$$w_n=\frac{2n+5}{n+1}.$$

By making use of Euclid’s division algorithm, we can rewrite the general term’s formula as

$$w_n=2+\frac{3}{n+1}.$$

Now, we observe that the sequence

$$\left(\frac{1}{n+1}\right)$$

is strictly decreasing, bounded above by its first term and below by 0, and we proceed as it follows:

$$\begin{array}{ll}& 0<\frac{1}{n+1}\le \frac{1}{2}\\ \iff & 0<\frac{3}{n+1}\le \frac{3}{2}\\ \iff & 2<2+\frac{3}{n+1}\le 2+\frac{3}{2},\end{array}$$

which allows us to conclude that $2<w_n\le \frac{7}{2}$, that is, $\left(w_n\right)$ is a bounded sequence.

2.2 Arithmetic progressions

In this section we will survey a class of sequences on which each term is obtained from its previous by adding a constant.

Definition 2.4 (Arithmetic progression). An arithmetic progression $\left(u_n\right)$ is a sequence defined by recurrence as

$$\left\{\begin{array}{lll}{u}_1& =& a\\ u_{n+1}& =& u_n+d,\forall n\in \mathbb{N},\end{array}\right.$$

where a and d are real constants.

The constant a in Definition 2.4 corresponds to the first term of the arithmetic progression. It is immediate to observe that the difference between any two consecutive terms of an arithmetic progression is constant and equal to d:

$$\forall n\in \mathbb{N},u_{n+1}-u_n=d$$

and, in view of this property, d is normally called the common difference of the arithmetic sequence $\left(u_n\right)$.

Example 2.9. Consider the sequence $\left(u_n\right)$ whose first terms are:

$$2,5,8,11,14,\dots$$

$\left(u_n\right)$ is the arithmetic sequence of first term 2 and common difference 3.

The value of the common difference of an arithmetic sequence contains all the information we need to analyze the sequence’s monotony. In fact, if the common difference is positive, that means that the arithmetic sequence is strictly increasing, since each term increases relatively to its predecessor by adding a positive constant. Similarly, if the value of the common difference is negative, then, by an analogous argument, the sequence is strictly decreasing. The particular case in which we have a null common difference yields a trivial constant arithmetic sequence.

One can easily derive a formula for the general term of an arithmetic sequence. Consider the arithmetic sequence of the first term a and common difference d. Then, we have

$$\begin{array}{lll}{u}_1& =& a,\\ u_2& =& u_1+d=a+d,\\ u_3& =& u_2+d=\left(k+d\right)+c=a+2d,\\ u_4& =& u_3+d=\left(k+2d\right)+c=a+3d,\\ & ⋮& \\ u_n& =& a+\left(n-1\right)d,\end{array}$$E4

which is the general term of the arithmetic sequence of first term a and common difference d.

Example 2.10. Continuing Example 2.9, the general term of the arithmetic sequence $\left(u_n\right)$, of first term 2 and common difference 3, is

$$u_n=2+\left(n-1\right)\times 3=2+3n-3=3n-1.$$

The general term formula (4) can be generalized for any given term of the sequence:

$$u_n=u_m+\left(n-m\right)d.$$

This means that we do not need the first term of the sequence to compute the general term. Any other term can do the trick.

Example 2.11. Continuing Examples 2.9 and 2.10, the general term of the arithmetic sequence $\left(u_n\right)$, of first term 2 and common difference 3, can also be computed by using the third term $u_3=8$:

$$u_n=8+\left(n-3\right)\times 3=8+3n-9=3n-1.$$

We will finish this section with a formula for calculating the sum of a finite number of terms of an arithmetic sequence. Let $\left(u_n\right)$ be an arithmetic sequence of common difference d. Then, we have

$$\begin{array}{lll}2\sum _{i=1}^n{u}_i& =& 2\left(u_1+u_2+u_3+\cdots +u_{n-2}+u_{n-1}+u_n\right)\\ & =& \left(u_1+u_n\right)+\left(u_2+u_{n-1}\right)+\left(u_3+u_{n-2}\right)+\cdots +\\ & +& \left(u_{n-2}+u_3\right)+\left(u_{n-1}+u_2\right)+\left(u_n+u_1\right)\end{array}$$E5

Observe that, for any natural m we have:

$$\begin{array}{lll}{u}_m+u_{n-\left(m-1\right)}& =& u_1+\left(m-1\right)d+u_1+\left(n-\left(m-1\right)-1\right)d\\ & =& u_1+\mathit{md}-d+u_1+\mathit{nd}-\mathit{md}\\ & =& u_1+u_1+\left(n-1\right)d\\ & =& u_1+u_n.\end{array}$$E6

Then, by making use of property (6), one can rewrite equality (5) as:

$$2\sum _{i=1}^n{u}_i=\left(u_1+u_n\right)\times n,$$E7

which, in turn, allows us to conclude that the sum, $S_n$, of the first n terms of an arithmetic sequence, $\left(u_n\right)$, regardless of its common difference, can be given by:

$$S_n=\sum _{i=1}^n{u}_i=\frac{u_1+u_n}{2}\times n.$$E8

This formula is usually associated with the German mathematician Carl Friedrich Gauss, who allegedly surprised the teacher by computing the sum of the first hundred natural numbers, 5050, by adding 1 to 100, multiplying the result by 100, and dividing by 2.

Formula (8) can be generalized in a straightforward way to calculate the sum of any consecutive terms of the sequence $\left(u_n\right)$ between terms $u_m$ and $u_n$ as:

$$S_{m,n}=\sum _{i=m}^n{u}_i=\frac{u_m+u_n}{2}\times \left(n-m+1\right).$$E9

Example 2.12. Continuing Examples 2.9, 2.10, and 2.11, the sum of the ten first terms of the arithmetic sequence $\left(u_n\right)$, of first term 2 and common difference 3, is:

$$S_{10}=\frac{2+3\times 10-1}{2}\times 10=\frac{310}{2}=155.$$

The sum of the ten consecutive terms of $\left(u_n\right)$ starting from $u_3$ is:

$$S_{3:12}=\frac{3\times 3-1+3\times 12-1}{2}\times 10=\frac{43}{2}\times 10=215.$$

2.3 Geometric progressions

In this section, a different type of sequence is presented. Contrary to the arithmetic progressions presented in Section 2.2, which have a linear growth, the sequences presented next have an exponential growth, as each of its terms is obtained by multiplying, not adding, its predecessor by a constant number.

Definition 2.5 (Geometric progression). A geometric progression $\left(u_n\right)$ is a sequence defined by recurrence as

$$\left\{\begin{array}{lll}{u}_1& =& a\\ u_{n+1}& =& u_n\times r,\forall n\in \mathbb{N},\end{array}\right.$$

where a and r are real constants.

The value a is the first term of the sequence. If $a=0$, then the geometric progression is the null sequence. Also, if $a\ne 0$ but $r=0$, then all the terms of the geometric progression from the second onwards will be null. In the remaining case, that is, with $a\ne 0$ and $r\ne 0$, all the terms of the geometric progression will not be null. In this latter case, as an immediate consequence, we observe that the ratio between any two consecutive terms of a geometric progression is constant and equal to r:

$$\forall n\in \mathbb{N},\frac{u_{n+1}}{u_n}=r,$$

and because of this r is usually referred to as the common ratio of the geometric progression.

Example 2.13. Consider the sequence $\left(u_n\right)$ of the numbers

$$3,6,12,24,48,\dots$$

The sequence $\left(u_n\right)$ is the geometric progression of first term 3 and common ratio 2.

As we did for the arithmetic progressions, we can also derive a formula for the general term of a geometric progression. Consider the geometric sequence of the first term a and common ratio r. Then, we have

$$\begin{array}{lll}{u}_1& =& a,\\ u_2& =& u_1\times r=a\times r,\\ u_3& =& u_2\times r=\left(a\times r\right)\times r=a\times r^2,\\ u_4& =& u_3\times r=\left(a\times r^2\right)\times r=a\times r^3,\\ & ⋮& \\ u_n& =& a\times r^{n-1},\end{array}$$E10

which is the general term of the geometric sequence of the first term a and common ratio r.

Example 2.14. Continuing Example 2.13, we have that the general term of the geometric progression $\left(u_n\right)$ of first term 3 and common ratio 2 is:

$$u_n=3\times 2^{n-1}.$$

One can obtain the general term’s formula by making use of any term of the geometric progression and not necessarily the first term. In fact, it is easy to obtain that:

$$u_n=u_m\times r^{n-m},$$E11

for any given term of order $m\in \mathbb{N}$.

Example 2.15. Continuing Examples 2.13 and 2.14, we can also obtain the general term of the geometric progression $\left(u_n\right)$ of first term 3 and common ratio 2, by using the term $u_3=12$ and formula (11):

$$u_n=12\times 2^{n-3}=3\times 2^2\times 2^{n-3}=3\times 2^{2+n-3}=3\times 2^{n-1}.$$

The analysis of the monotony of geometric progressions is not as straightforward as in the case of the arithmetic progressions. Let’s exclude the trivial cases with first term or common ratio equal to 0 which yield null sequences at least from the second term onwards.

Proposition 2.1. Let $\left(u_n\right)$ be a geometric progression with $u_1\ne 0$ and common ration $r\ne 0$. The following hold:

1. If $r=1$, then $\left(u_n\right)$ is constant and all terms are equal to $u_1$.

2. If $r>1$, then

   1. if $u_1>0$, then $\left(u_n\right)$ is increasing.

   2. if $u_1<0$, then $\left(u_n\right)$ is decreasing.

3. If $0<r<1$, then

   1. if $u_1>0$, then $\left(u_n\right)$ is decreasing.

   2. if $u_1<0$, then $\left(u_n\right)$ is increasing.

4. If $r<0$, then $\left(u_n\right)$ is not monotonic, as it has alternating sign terms.

The proof of Proposition 2.1 is left as an exercise for the reader.

As in the previous section, there is also a formula for calculating the sum, $S_n$, of n consecutive terms of a geometric progression $\left(u_n\right).$

The case where $r=1$ is trivial. Since, in this case, all terms are equal to $u_1$, then $S_n=n\times u_1$.

Let’s consider the more general case with $r\ne 1$. We have that

$$S_n=\sum _{i=1}^n{u}_i=u_1+u_2+u_3+\cdots +u_{n-1}+u_n.$$E12

Multiplying both members of equality (12) by r, we obtain

$$\begin{array}{lll}{S}_{n}r& =& u_{1}r+u_{2}r+u_{3}r+\cdots +u_{n-1}r+u_{n}r\\ & =& u_2+u_3+u_4+\cdots +u_n+u_{n+1}.\end{array}$$E13

Now, subtracting equalities (12) and (13), we obtain:

$$\begin{array}{ll}& S_n-S_{n}r=u_1+u_2+u_3+\cdots +u_{n-1}+u_n\\ & -\left(u_2+u_3+u_4+\cdots +u_n+u_{n+1}\right)\\ \iff & S_n\left(1-r\right)=u_1-u_{n+1}\\ \iff & S_n\left(1-r\right)=u_1-u_{n}r\\ \iff & S_n\left(1-r\right)=u_1-u_1{r}^{n-1}r\\ \iff & S_n\left(1-r\right)=u_1-u_1{r}^n\\ \iff & S_n\left(1-r\right)=u_1\left(1-r^n\right)\\ \iff & S_n=\frac{u_1\left(1-r^n\right)}{1-r},\end{array}$$E14

which gives the sum of the n first terms of the geometric progression $\left(u_n\right)$ with first term $u_1$ and common ratio r. Note that formula (14) can be easily applied to calculate the sum of any consecutive terms of $\left(u_n\right)$ between $u_m$ and $u_n$:

$$S_{m:n}=\frac{u_m\left(1-r^{n-m+1}\right)}{1-r}.$$E15

Example 2.16. Continuing Examples 2.13, 2.14, and 2.15, we have that the sum the ten consecutive terms of the geometric progression $\left(u_n\right)$ of first term 3 and common ratio 2, starting at term $u_4=24$ is:

$$S_{4:13}=\frac{24\left(1-2^{10}\right)}{1-2}=24552.$$

2.4 Convergent sequences

Some sequences have an interesting property: their terms approximate to a certain value as the order increases. This kind of sequence is called convergent, and we will survey it along this section.

Definition 2.6 (Limit of a sequence, convergent sequence). Let $\left(u_n\right)$ be a sequence. A real number a is said to be the limit of $\left(u_n\right)$ if

$$\forall \epsilon >0,\exists m\in \mathbb{N}:\forall n\in \mathbb{N},n\ge m\Rightarrow \mid u_n-a\mid <\epsilon ,$$

and we write ${lim}_{n\to +\infty }{u}_n=a,{lim}_n{u}_n=a$ or simply $lim{u}_n=a$. If a number a exists in these conditions, $\left(u_n\right)$ is said to be convergent. Otherwise, $\left(u_n\right)$ is divergent.

When we say that $lim{u}_n=a$ the idea is that we can always find an order from which all the terms are as close as we want, or as we previously defined, from the value a. Therefore, $\left(u_n\right)$ gets closer and closer to that value a, that is, it converges to a.

Example 2.17. Consider the sequence $\left(u_n\right)$ of general term

$$u_n=\frac{2n+1}{n}.$$

Let’s prove that $lim{u}_n=2$.

Let $\epsilon >0$. We have

$$\begin{array}{lll}\left|u_n-2\right|<\epsilon & \iff & \left|\frac{2n+1}{n}-2\right|<\epsilon \\ & \iff & \left|\frac{2n+1-2n}{n}\right|<\epsilon \\ & \iff & \left|\frac{1}{n}\right|<\epsilon \\ & \iff & \frac{1}{n}<\epsilon \\ & \iff & n>\frac{1}{\epsilon }.\end{array}$$

Then, for all $\epsilon >0$, there is a natural number m such that for all $n\in \mathbb{N}$, we have $n\ge m\Rightarrow \mid u_n-a\mid <\epsilon$. That number m can be chosen as any natural number bigger that $1/\epsilon$. Hence, $lim{u}_n=2$.

An expected property of the definition of limit of a sequence is that when the limit exists, it is unique; that is, a convergent sequence cannot have two distinct limits.

There are two particular cases of divergent sequences that should be highlighted.

Definition 2.7 (Infinite limit of a sequence). Let $\left(u_n\right)$ be a sequence.

1. It is said that $\left(u_n\right)$ has limit $+\infty$, and we write $lim{u}_n=+\infty$, if

   $$\forall L>0,\exists m\in \mathbb{N}:\forall n\in \mathbb{N},n\ge m\Rightarrow u_n>L.$$

2. It is said that $\left(u_n\right)$ has limit $-\infty$, and we write $lim{u}_n=-\infty$, if

   $$\forall L>0,\exists m\in \mathbb{N}:\forall n\in \mathbb{N},n\ge m\Rightarrow u_n<-L.$$

Example 2.18. Let $\left(u_n\right)$ and $\left(v_n\right)$ be the sequences defined by $u_n=3n+2$ and $v_n=3-2n,$ respectively. Let’s prove that $lim{u}_n=+\infty$ and $lim{v}_n=-\infty$.

Let $L>0$.

Regarding $\left(u_n\right)$, we have

$$\begin{array}{lll}{u}_n>L& \iff & 3n+2>L\\ & \iff & n>\frac{L-2}{3}.\end{array}$$

Then, there is a order $m\in \mathbb{N}$ such that, for all $n\in \mathbb{N}$, we have

$$n\ge m\Rightarrow u_n>L.$$

We just have to choose m to be a natural number bigger that $\left(L-2\right)/3$. Hence, $lim{u}_n=+\infty$.

Regarding $\left(v_n\right)$, we have

$$\begin{array}{lll}{v}_n<-L& \iff & 3-2n<-L\\ & \iff & n>\frac{L+3}{2}.\end{array}$$

Then, there is a order $m\in \mathbb{N}$ such that, for all $n\in \mathbb{N}$, we have

$$n\ge m\Rightarrow v_n<-L.$$

In this case, it is enough to choose m to be a natural number bigger that $\left(L+3\right)/2$. Hence, $lim{v}_n=-\infty$.

There are some interesting and well-known results involving the convergence of sequences.

Theorem 2.1. Let X be a finite subset of $\mathbb{N}$ and $\left(u_n\right)$ and $\left(v_n\right)$ be two sequences such that, for all $n\in \mathbb{N}\backslash X$, we have $u_n=v_n$. Then, $\left(u_n\right)$ is convergent if and only if $\left(v_n\right)$ is convergent. Also, $lim{u}_n=lim{v}_n$.

Theorem 2.1 basically states that two different sequences that differ only in a finite number of terms converge or diverge in the same manner. We leave the proof as an exercise for the reader.

Example 2.19. Let $\left(u_n\right)$ be the sequence whose general term is $u_n=2-\sqrt{n}$ and $\left(v_n\right)$ the sequence defined by:

$$v_n=\left\{\begin{array}{ll}\frac{1}{n}+2,& \mathit{\text{if}}n<25\\ u_n,& \mathit{\text{if}}n\ge 25\end{array}\right..$$

Since $\left(u_n\right)$ and $\left(v_n\right)$ only differ in a finite number of terms (the first 24 terms), then $lim{v}_n=lim{u}_n=2-\infty =-\infty$.

As it is implied in the resolution of Example 2.19, let it be clear that the first step in evaluating a sequence’s limit is to replace every instance of n, in the sequence’s general term, with $+\infty$ and calculate. Operating with $\infty$ is, therefore, of great importance. We will not elaborate on this topic for now, as this section deals with a more theoretical part of convergence, but it will be one of the main topics surveyed in Section 2.5.

Theorem 2.2. Let $\left(u_n\right)$ be a sequence. If $\left(u_n\right)$ is convergent, then $\left(u_n\right)$ is bounded.

Proof. Let $\left(u_n\right)$ be a convergent sequence and suppose that $lim{u}_n=a.$ Then, by definition, for any fixed $\epsilon >0$, there is an order $m\in \mathbb{N}$ such that $\mid u_n-a\mid <\epsilon ,$ for all $n\ge m$. This means that the set of terms of $\left(u_n\right)$ with $n>m$ belongs to ${\mathcal{V}}_a$, the neighborhood of a. We observe that $a+\epsilon$ and $a-\epsilon$ constitute an upper bound and a lower bound of that set of terms of $\left(u_n\right)$, respectively; thus, that set is bounded.

As for the remaining terms of $\left(u_n\right),$ that is, the terms with $n<m$ they constitute a finite set that is naturally bounded. We can, therefore, conclude that all terms of $\left(u_n\right)$ are bounded, hence $\left(u_n\right)$ is a bounded sequence. □

Observe that the converse from Theorem 2.2 is not true. Consider the trivial alternating sequence, that is, a sequence whose terms alternate between positive and negative, with general term ${\left(-1\right)}^n$. It is clearly a bounded sequence, but it does not converge as it keeps alternating from −1 to 1.

Theorem 2.3 (Squeeze Theorem for Convergent Sequences). Let $\left(u_n\right),\left(v_n\right)$ and $\left(w_n\right)$ be sequences such that

$$\exists m\in \mathbb{N}:\forall n\in \mathbb{N},n>m\Rightarrow u_n\le v_n\le w_n.$$

If $lim{u}_n=lim{w}_n=a$, where $a\in \mathbb{R}$, then $lim{v}_n=a$.

Proof. Let $\epsilon >0$. Then, since $lim{u}_n=lim{w}_n=a$, there are natural numbers $m_1$ and $m_2$ such that

$$\begin{array}{l}n\ge m_1\Rightarrow \mid u_n-a\mid <\epsilon \iff a-\epsilon <u_n<a+\epsilon ,\\ n\ge m_2\Rightarrow \mid w_n-a\mid <\epsilon \iff a-\epsilon <w_n<a+\epsilon .\end{array}$$

Then, considering $m=max\left\{m_1,m_2\right\}$, we have

$$n\ge m\Rightarrow a-\epsilon <u_n\le v_n\le w_n<a+\epsilon ,$$

which means that

$$n\ge m\Rightarrow \mid v_n-a\mid <\epsilon .$$

Hence, $lim{v}_n=a$. □

Note that the result of Theorem 2.3 can be straightforwardly extended to the cases where $a=\pm \infty$.

Example 2.20. Let $\left(u_n\right)$ be the sequence whose general term is

$$u_n=\frac{sinn}{n}.$$

Since the sine yields values in the interval [−1, 1], we can write

$$-\frac{1}{n}\le \frac{sinn}{n}\le \frac{1}{n}.$$

Now, observing that $lim\left(-\frac{1}{n}\right)=lim\frac{1}{n}=0$, we can conclude, by Theorem 2.3, that $lim{u}_n=0$.

From Theorem 2.3, one can prove the following result.

Theorem 2.4. Let $\left(u_n\right)$ be a sequence such that $lim\mid u_n\mid =0$. Then, $lim{u}_n=0$.

Theorem 2.4 is especially useful for computing certain limits involving alternating sequences.

Example 2.21. Let $\left(u_n\right)$ be the sequence defined by

$$u_n=\frac{{\left(-1\right)}^n}{2n+1}.$$

Since

$$lim\mid u_n\mid =lim\frac{1}{2n+1}=\frac{1}{+\infty }=0,$$

we can conclude, by Theorem 2.4, that $lim{u}_n=0$.

Theorem 2.5. Let $\left(u_n\right)$ and $\left(v_n\right)$ be two convergent sequences such that

$$lim{u}_n<lim{v}_n.$$

Then, there is a $m\in \mathbb{N}$ such that

$$n\ge m\Rightarrow u_n<v_n.$$

Proof. Let $\left(u_n\right)$ and $\left(v_n\right)$ be two convergent sequences with $lim{u}_n=a$ and $lim{v}_n=b$ such that $a<b$. Let $\epsilon =\frac{b-a}{2}$. Note that . By the definition of limit, there are natural numbers $m_1$ and $m_2$ such that

$$\begin{array}{l}n\ge m_1\Rightarrow \mid u_n-a\mid <\epsilon \iff a-\epsilon <u_n<a+\epsilon ,\\ n\ge m_2\Rightarrow \mid v_n-b\mid <\epsilon \iff b-\epsilon <v_n<b+\epsilon .\end{array}$$

Now, observing that $a+\epsilon =b-\epsilon =\frac{a+b}{2}$, and taking $m=max\left\{m_1,m_2\right\},$ we have that

$$n>m\Rightarrow u_n<a+\epsilon =b-\epsilon <v_n.$$

□

Theorem 2.6. Let $\left(u_n\right)$ be a sequence.

1. If $\left(u_n\right)$ is increasing and bounded from above, then it is convergent.

2. If $\left(u_n\right)$ is decreasing and bounded from below, then it is convergent.

3. If $\left(u_n\right)$ is monotonic and bounded, then it is convergent.

Proof. We will just present proof for assertion 1 from Theorem 2.6. Assertion 2 is analogous, and assertion 3 is a direct consequence from the previous two.

Let $\left(u_n\right)$ be an increasing sequence bounded from above. We can immediately conclude that $\left(u_n\right)$ is bounded, that is, that the set of terms of $\left(u_n\right)$ is a bounded set. Since it is bounded and it is a non-empty set, then we can conclude that it has a supremum. Let a be the supremum of the set of terms of $\left(u_n\right)$. We will prove that $lim{u}_n=a$.

Let $\epsilon >0$. By definition of supremum, we have that, for all $n\in \mathbb{N},u_n\le a$. We also have that there is at least one $m\in \mathbb{N}$ such that $u_m>a-\epsilon .$ Since $\left(u_n\right)$ is an increasing sequence, we also must have that, for all $n>m,u_n\ge u_m>a-\epsilon .$ Therefore, we can conclude that there is an order $m\in \mathbb{N}$ such that, for all $n\in \mathbb{N}$, we have $\mid u_n-a\mid <\epsilon ,$ if $n>m$. Thus, $lim{u}_n=a$ and $\left(u_n\right)$ is a convergent sequence. □

The result presented above has direct application as it is shown in the following example.

Example 2.22. Consider the sequence $\left(u_n\right)$ with general term

$$u_n=\frac{2n-1}{n}.$$

Let’s show that $\left(u_n\right)$ is convergent.

Firstly, since

$$\begin{array}{lll}{u}_{n+1}-u_n& =& \frac{2\left(n+1\right)-1}{n+1}-\frac{2n-1}{n}\\ & =& \frac{2n+1}{n+1}-\frac{2n-1}{n}\\ & =& \frac{\left(2n+1\right)n-\left(2n-1\right)\left(n+1\right)}{n\left(n+1\right)}\\ & =& \frac{1}{n\left(n+1\right)}\\ & >& 0,\end{array}$$

then $\left(u_n\right)$ is increasing.

Now, observe that $u_n=2-1/n$ and that

$$\begin{array}{ll}& 0<\frac{1}{n}\le 1\\ \iff & -1\le -\frac{1}{n}<0\\ \iff & 1\le 2-\frac{1}{n}<2.\end{array}$$

Hence, $\left(u_n\right)$ is bounded. Then, from Theorem 2.6, we can conclude that $\left(u_n\right)$ is convergent.

Note that being convergent does not imply being monotonic. For instance, the sequence with general term ${\left(-1\right)}^n/n$ is convergent to 0, but it is not monotonic as it is an alternating sequence.

Theorem 2.7. Let $\left(u_n\right)$ and $\left(v_n\right)$ be two sequences such that $u_n$ is bounded and $lim{v}_n=0$. Then $lim\left(u_n\times v_n\right)=0.$

Proof. Let $\left(u_n\right)$ be a bounded sequence and $\left(v_n\right)$ be a sequence such that $lim{v}_n=0$. Let $w_n=u_n\times v_n$ and $\epsilon >0$.

Since $\left(u_n\right)$ is bounded, then there is a $k\in \mathbb{R}$ such that $\mid u_n\mid \le k$, for all $n\in \mathbb{N}$. Then, we can write

$$\mid w_n\mid =\mid u_n\times v_n\mid =\mid u_n\mid \times \mid v_n\mid \le k\mid v_n\mid .$$

Now, since $lim{v}_n=0$, there is an order $m\in \mathbb{N}$ such that, for all $n>m,v_n<\epsilon /k.$ Then, we can conclude that, for all $n>m$, we have

$$\mid w_n\mid \le k\mid v_n\mid <\epsilon ,$$

that is, $lim{w}_n=0.$ □

Some potentially complicated limits can be computed using the result presented in Theorem 2.7.

Example 2.23. Let’s compute the value of

$$lim\frac{cos\left(n-2\right)}{n+1}.$$

Observe that

$$\frac{cos\left(n-2\right)}{n+1}=cos\left(n-2\right)\times \frac{1}{n+1}.$$

The sequence whose general term is $cos\left(n-2\right)$ is clearly bounded, as the cosine yields values in the interval [−1, 1].

As for the sequence whose general term is $1/\left(n+1\right)$, we have that

$$lim\frac{1}{n+1}=\frac{1}{+\infty }=0.$$

Then, we can conclude, by Theorem 2.7, that

$$lim\frac{cos\left(n-2\right)}{n+1}=0$$

In the following, we present a set of operating properties regarding limits of sequences.

Proposition 2.2. Let $\left(u_n\right)$ and $\left(v_n\right)$ be convergent sequences such that $lim{u}_n=a$ and $lim{v}_n=b.$ Let $\alpha \in \mathbb{R}$ and $k\in \mathbb{N}$. We have:

1. $lim\left(u_n+v_n\right)=a+b;$

2. $lim\left(u_n-v_n\right)=a-b;$

3. $lim\left(u_n\times v_n\right)=\mathit{ab};$

4. $lim\left(\alpha u_n\right)=\mathit{\alpha a};$

5. $lim{\left(u_n\right)}^k=a^k;$

6. $lim\mid u_n\mid =\mid a\mid ;$

7. $lim\left(\frac{1}{u_n}\right)=\frac{1}{a}$, if $u_n\ne 0,\forall n\in \mathbb{N}$ and $a\ne 0;$

8. $lim\left(\frac{u_n}{v_n}\right)=\frac{a}{b}$, if $v_n\ne 0,\forall n\in \mathbb{N}$ and $b\ne 0;$

9. $lim\sqrt[k]{u_n}=\sqrt[k]{a}$, with if $u_n\ge 0,\forall n\in \mathbb{N}$, if k is an even number.

We finish this survey on sequences by presenting a very important class of sequences called the Cauchy sequences, named after the French mathematician Augustin-Louis Cauchy. Their relevance and their implications on convergence make them frequently referred to as fundamental sequences.

Definition 2.8 (Cauchy sequence). The sequence $\left(u_n\right)$ is a Cauchy sequence if

$$\forall \epsilon >0,\exists p\in \mathbb{N}:\forall m,n\in \mathbb{N},m,n>p\Rightarrow \mid u_m-u_n\mid <\mathrm{ϵ}.$$

Observe that the above formal definition simply states that the terms of a Cauchy sequence become as close as we want as the sequence progresses.

Theorem 2.8. Let $\left(u_n\right)$ be a Cauchy sequence. Then $\left(u_n\right)$ is bounded.

Proof. Let $\left(u_n\right)$ be a Cauchy sequence. By definition, for any fixed $\epsilon >0$, there is an order $p\in \mathbb{N}$ such that $\mid u_m-u_n\mid <\epsilon ,$ for all natural numbers $m,n>p$. In particular, if $n=p+1$, we have that $\mid u_m-u_{p+1}\mid <\epsilon ,$ for all $m>p$, which means that the set of terms of $\left(u_n\right)$ with $n>p$ is bounded as $-\epsilon +u_{p+1}$ is a lower bound and $\epsilon +u_{p+1}$ is an upper bound for this set. As for the set of terms of $\left(u_n\right)$ with $n<p$, since it is a finite set, it is also bounded. Hence, $\left(u_n\right)$ is a bounded sequence.

□

Using Theorem 2.8, one can prove the following result that states that being a convergent sequence and being a Cauchy sequence are the same thing.

Theorem 2.9. Let $\left(u_n\right)$ be a sequence. Then, $\left(u_n\right)$ is convergent if and only if $\left(u_n\right)$ is a Cauchy sequence.

2.5 Techniques for evaluating limits

In this section, we will present a set of relevant tools for computing the limits of certain sequences. We will also analyze the main indeterminate forms that one can attain when computing limits, and we will discuss techniques to evaluate these indeterminate forms.

We will start this section with a result that can be of extreme importance in all domains of sequence analysis.

Theorem 2.10. Let $\left(u_n\right)$ be a sequence and f a real function such that, $\forall n\in \mathbb{N},f\left(n\right)=u_n$. Then we have:

1. If $\underset{x\to +\infty }{lim}f\left(x\right)=a$, for $a\in \mathbb{R}$, then $lim{u}_n=a$.

2. If f is increasing (decreasing), then $\left(u_n\right)$ is increasing (decreasing).

This result allows us to apply functional analysis to solve some problems involving sequences by considering a corresponding real function.

Example 2.24. Calculating the limit of the sequence $\left(u_n\right)$ defined by

$$u_n=\frac{3n}{ln\left(n\right)+2n}$$

may prove to be difficult at first glance, but if one considers the real function f defined by

$$f\left(x\right)=\frac{3x}{ln\left(x\right)+2x},$$

then we obtain, by applying the well-known L’Hôpital’s Rule,

$$\begin{array}{lll}\underset{x\to +\infty }{lim}f\left(x\right)& =& \underset{x\to +\infty }{lim}\frac{3x}{ln\left(x\right)+2x}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& \underset{x\to +\infty }{lim}\frac{\frac{d}{\mathit{dx}}\left(3x\right)}{\frac{d}{\mathit{dx}}\left(ln\left(x\right)+2x\right)}\\ & =& \underset{x\to +\infty }{lim}\frac{3}{\frac{1}{x}+2}\\ & =& \frac{3}{0+2}\\ & =& \frac{3}{2}.\end{array}$$

By Theorem 2.10, we can conclude that $lim{u}_n=3/2$.

Example 2.24 brings out again the need to clarify operations involving $\infty$. The following proposition presents a list of relevant properties. Before, however, let us introduce the following notation.

Let $\left(u_n\right)$ be a convergent sequence with limit equal to a. If $\left(u_n\right)$ converges to the value a through values which are strictly smaller than a, then we write $lim{u}_n=a^{-}$. If $\left(u_n\right)$ converges to the value a through values which are strictly bigger than a, then we write $lim{u}_n=a^{+}$.

As a trivial example, consider the sequences $\left(u_n\right)$ and $\left(v_n\right)$ whose general terms are $u_n=1/n$ and $v_n=-1/n.$ It is easy to verify that $lim{u}_n=0^{+}$ and $lim{v}_n=0^{-}$.

Proposition 2.3. While evaluating a limit, the following equalities hold.

1. $a\pm \infty =\pm \infty ,\forall a\in \mathbb{R};$

2. $+\infty +\infty =+\infty ;$

3. $-\infty -\infty =-\infty ;$

4. $a\times \left(\pm \infty \right)=\pm \infty ,\forall a\in {\mathbb{R}}^{+}$ or $a=+\infty ;$

5. $a\times \left(\pm \infty \right)=\mp \infty ,\forall a\in {\mathbb{R}}^{-}$ or $a=-\infty ;$

6. $\frac{a}{\pm \infty }=0,\forall a\in \mathbb{R};$

   1. $\frac{a}{+\infty }=0^{+},\forall a\in {\mathbb{R}}^{+};$

   2. $\frac{a}{-\infty }=0^{-},\forall a\in {\mathbb{R}}^{+};$

   3. $\frac{a}{+\infty }=0^{-},\forall a\in {\mathbb{R}}^{-};$

   4. $\frac{a}{-\infty }=0^{+},\forall a\in {\mathbb{R}}^{-};$

7. $\frac{a}{0}=\infty ,\forall a\in \mathbb{R}\backslash \left\{0\right\};$

   1. $\frac{a}{0^{+}}=+\infty ,\forall a\in {\mathbb{R}}^{+}$ or $a=+\infty ;$

   2. $\frac{a}{0^{+}}=-\infty ,\forall a\in {\mathbb{R}}^{-}$ or $a=-\infty ;$

   3. $\frac{a}{0^{-}}=-\infty ,\forall a\in {\mathbb{R}}^{+}$ or $a=+\infty ;$

   4. $\frac{a}{0^{-}}=+\infty ,\forall a\in {\mathbb{R}}^{-}$ or $a=-\infty ;$

8. ${\left(\infty \right)}^k=\infty ,\forall k\in \mathbb{N};$

   1. ${\left(+\infty \right)}^k=+\infty ,\forall k\in \mathbb{N};$

   2. ${\left(-\infty \right)}^k=+\infty ,\forall k\in \mathbb{N}$ and k even;

   3. ${\left(-\infty \right)}^k=-\infty ,\forall k\in \mathbb{N}$ and k odd;

9. $\sqrt[k]{+\infty }=+\infty ,\forall k\in \mathbb{N}\backslash \left\{1\right\};$

10. $\sqrt[k]{-\infty }=-\infty ,\forall k\in \mathbb{N}\backslash \left\{1\right\}$ and k is odd.

The properties presented are invaluable to operate with $\infty$ and compute the values of limits by direct substitution of the variable n in the general term by $+\infty$. The following example presents some cases.

Example 2.25. Let’s compute the value of the following limits.

1. $lim\left(2n-5\right)$

   $$\begin{array}{lll}lim\left(2n-5\right)& =& 2\times \left(+\infty \right)-5\\ & =& +\infty -5\\ & =& +\infty .\end{array}$$

2. $lim\frac{2}{3-3n}$

   $$\begin{array}{lll}lim\frac{2}{3-3n}& =& \frac{2}{3-3\times \left(+\infty \right)}\\ & =& \frac{2}{3-\infty }\\ & =& \frac{2}{-\infty }\\ & =& 0.\end{array}$$

3. $lim\left[n^2\left(1-n\right)\right]$

   $$\begin{array}{lll}lim\left[n^2\left(1-n\right)\right]& =& {\left(+\infty \right)}^2\times \left[1-\left(+\infty \right)\right]\\ & =& +\infty \times \left(1-\infty \right)\\ & =& +\infty \times \left(-\infty \right)\\ & =& -\infty .\end{array}$$

Left out from the list presented in Proposition 2.3 are the cases of $+\infty -\infty ,\frac{\infty }{\infty },\frac{0}{0},0\times \infty ,{\infty }^0,0^0$ and $1^{\infty }$. These are called indeterminate forms as they can actually be equal to different real values or $\pm \infty$ depending on the situation at hand. We will observe this particular behavior in the following paragraphs, on which we will survey some of the techniques one can apply to evaluate different types of indeterminate forms.

2.5.1 Indeterminate form $\infty -\infty$

This type of indeterminate form, $\infty -\infty$, usually arises from evaluating the limit of sequences constructed in the form of a polynomial of variable n, although it can also appear in other contexts. For example, if one attempts to evaluate the limit of the sequence $\left(u_n\right)$ whose general term is $u_n=n^2-2n+1$ by direct substitution, one will obtain this indeterminate form.

In the context of sequences defined by polynomials, this difficulty is very easy to surpass. Indeed, one needs only to collect the higher degree term in $\left(u_n\right)$ and substitute again. The indeterminate form will disappear.

Example 2.26. Consider the sequence $\left(u_n\right)$ with general term $u_n=n^2-2n+1$. We have:

$$\begin{array}{lll}lim{u}_n& =& lim\left(n^2-2n+1\right)\\ & \stackrel{\left(\infty -\infty \right)}{=}& n^2\left(1-\frac{2}{n}+\frac{1}{n^2}\right)\\ & =& +\infty \left(1-\frac{2}{+\infty }+\frac{1}{+\infty }\right)\\ & =& +\infty \left(1-0+0\right)\\ & =& +\infty .\end{array}$$

Example 2.26 leads to a more general property.

Proposition 2.4. Let p(x) be a non-null polynomial with leading coefficient a and $\left(u_n\right)$ be a sequence such that, $\forall n\in \mathbb{N},u_n=p\left(n\right)$. Then, the following statements hold.

1. If $a>0$, then $lim{u}_n=+\infty ;$

2. If $a<0$, then $lim{u}_n=-\infty$.

When defining sequences with the help of the concept of a polynomial, it is implied that the degree of the powers of n are greater or equal than 0. The next result lists clarifies all possibilities regarding the limits of rational powers of n.

Proposition 2.5. Let $q\in \mathbb{Q}$ and $\left(u_n\right)$ be the sequence whose general term is $u_n=n^q$. Then, the following hold.

1. If $q>0$, then $lim{u}_n=+\infty ;$

2. If $q=0$, then $lim{u}_n=1;$

3. If $q<0$, then $lim{u}_n=0$.

Proof. Let $q\in \mathbb{Q}$.

1. If $q>0$, let $L>0$. Then, we have

   $$\begin{array}{lll}{u}_n>L& \iff & n^q>L\\ & \iff & n>\sqrt[q]{L}.\end{array}$$

   Therefore, choosing any natural m bigger than $\sqrt[q]{L}$, we have that

   $$\forall n\in \mathbb{N},n\ge m\Rightarrow u_n>L,$$

   which is the same as saying that $lim{u}_n=+\infty$.

2. If $q=0$, then $u_n=1,\forall n\in \mathbb{N}$. Hence, $lim{u}_n=1$.

3. If $q<0$, let $\epsilon >0$. Then, we have

$$\begin{array}{lll}\mid u_n-0\mid <\epsilon & \iff & \mid n^q\mid <\epsilon \\ & \iff & \left|\frac{1}{n^{-q}}\right|<\epsilon \\ & \iff & \frac{1}{n^{-q}}<\epsilon \\ & \iff & n^{-q}>\frac{1}{\epsilon }\\ & \iff & n>\sqrt[-q]{\epsilon }.\end{array}$$

Therefore, choosing any natural m bigger than $\sqrt[-q]{\epsilon }$, we have that

$$\forall n\in \mathbb{N},n\ge m\Rightarrow \mid u_n-0\mid >\epsilon ,$$

which is the same as saying that $lim{u}_n=0$.

□

The next example presents another occasion where the indeterminate form appears.

Example 2.27. Consider the sequence $\left(u_n\right)$ with general term $u_n=\sqrt{n+1}-\sqrt{n}$. The limit of $\left(u_n\right)$ can be obtained by making use of the well-known multiplication identity $\left(A-B\right)\left(A+B\right)=A^2-B^2$ in the following manner:

$$\begin{array}{lll}lim{u}_n& =& lim\left(\sqrt{n+1}-\sqrt{n}\right)\\ & \stackrel{\left(\infty -\infty \right)}{=}& lim\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}\\ & =& lim\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\\ & =& lim\frac{1}{\sqrt{n+1}+\sqrt{n}}\\ & =& \frac{1}{+\infty +\infty }\\ & =& \frac{1}{+\infty }\\ & =& 0.\end{array}$$

We will finish this subsection by presenting one type of sequence for which the limit evaluation might result in indeterminate forms, particularly the one that is under the scope in these paragraphs.

Proposition 2.6. Let $a\in {\mathbb{R}}^{+}$ and $\left(u_n\right)$ be the sequence whose general term is $u_n=a^n$. Then, the following statements hold.

1. If $a>1$, then $lim{u}_n=+\infty ;$

2. If $a=1$, then $lim{u}_n=1;$

3. If $0<a<1$, then $lim{u}_n=0$.

Proof. Let $a\in {\mathbb{R}}^{+}$.

1. If $a>1$, let $L>0$. We have

   $$\begin{array}{lll}{u}_n>L& \iff & a^n>L\\ & \iff & n>{log}_{a}L.\end{array}$$

   Then, choosing any natural number bigger than ${log}_{a}L$, we have that

   $$\forall n\in \mathbb{N},n\ge m\Rightarrow u_n>L,$$

   which means that $lim{u}_n=+\infty$.

2. If $a=1$, then all terms of $u_n$ are equal to 1. Hence, $lim{u}_n=1$

3. If $0<a<1$, then $a^{-1}>1$, which allows us to write

   $$\begin{array}{lll}lim{u}_n& =& lim{a}^n\\ & =& lim\frac{1}{{\left(a^{-1}\right)}^n}\\ & =& \frac{1}{lim{\left(a^{-1}\right)}^n}\\ & =& \frac{1}{+\infty }\\ & =& 0.\end{array}$$

□

In the following, we present an example of an indeterminate form $\infty -\infty$ that arises from subtracting two sequences of the form presented in Proposition 2.6.

Example 2.28. Let $\left(u_n\right)$ be the sequence defined by $u_n=5^n-2^n$.

Computing the limit of $\left(u_n\right)$ gives us, by Proposition 2.6, the indeterminate form $\infty -\infty$. We can evaluate this limit by rewriting the general term, collecting the biggest power, and applying Proposition 2.6:

$$\begin{array}{lll}lim{u}_n& =& lim\left(5^n-2^n\right)\\ & \stackrel{\left(\infty -\infty \right)}{=}& lim\left[5^n\left(1-\frac{2^n}{5^n}\right)\right]\\ & =& lim\left[5^n\left(1-{\left(\frac{2}{5}\right)}^n\right)\right]\\ & =& +\infty \left(1-0\right)\\ & =& +\infty .\end{array}$$

2.5.2 Indeterminate form $\frac{\infty }{\infty }$

The type of indeterminate form represented by $\frac{\infty }{\infty }$ can appear when working out limits of sequences whose general term is a division of polynomials in the variable n, but it is not restricted to these cases. Trivial examples, which can be solved just by simplifying the fraction, are presented next.

Example 2.29. Let $\left(u_n\right),\left(v_n\right)$ and $\left(w_n\right)$ be the sequences whose general terms are

$$u_n=\frac{n^2}{3n},v_n=\frac{n}{3n}\mathit{\text{and}}{w}_n=\frac{n}{3n^2},$$

respectively.

By simplifying the general terms, we obtain

$$\begin{array}{lll}lim{u}_n& =& lim\frac{n^2}{3n}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{n}{3}\\ & =& \frac{+\infty }{3}\\ & =& +\infty ,\end{array}$$

$$\begin{array}{lll}lim{v}_n& =& lim\frac{n}{3n}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{1}{3}\\ & =& \frac{1}{3},\end{array}$$

$$\begin{array}{lll}lim{w}_n& =& lim\frac{n}{3n^2}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{1}{3n}\\ & =& \frac{1}{3\times \left(+\infty \right)}\\ & =& \frac{1}{+\infty }\\ & =& 0.\end{array}$$

Example 2.29 enlightens in a very simple way the concept of indeterminate form. Observe how the three cases of the indeterminate form $\frac{\infty }{\infty }$, after simplification, yield three different outcomes.

Some cases, however, cannot be solved through direct simplification only. Similarly to what we did in the previous section, collecting the higher degree terms of polynomials can be useful before simplifying. In the following example, three cases are presented using this technique.

Example 2.30. Consider the sequence $\left(u_n\right),\left(v_n\right)$ and $\left(w_n\right)$ with general terms defined by

$$u_n=\frac{2n^3-n^2+1}{n-3},v_n=\frac{3n^2+n-2}{-n^2-3n+1}\mathit{\text{and}}{w}_n=\frac{3-2n}{n^2+n+1},$$

respectively.

In each case, direct substitution ends with the indeterminate form $\frac{\infty }{\infty }$. The idea is to collect the term with a higher degree in the fraction’s numerator and repeat the process with the denominator. Then, simplify and, finally, substitute. We have:

$$\begin{array}{lll}lim{u}_n& =& lim\frac{2n^3-n^2+1}{n-3}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{2n^3\left(1-\frac{n^2}{2n^3}+\frac{1}{2n^3}\right)}{n\left(1-\frac{3}{n}\right)}\\ & =& lim\frac{2n^2\left(1-\frac{1}{2n}+\frac{1}{2n^3}\right)}{1-\frac{3}{n}}\\ & =& \frac{+\infty \left(1-\frac{1}{+\infty }+\frac{1}{+\infty }\right)}{1-\frac{3}{+\infty }}\\ & =& \frac{+\infty \left(1-0+0\right)}{1-0}\\ & =& +\infty ,\end{array}$$

$$\begin{array}{lll}lim{v}_n& =& lim\frac{3n^2+n-2}{-n^2-3n+1}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{3n^2\left(1+\frac{n}{3n^2}-\frac{2}{3n^2}\right)}{-n^2\left(1-\frac{3n}{-n^2}+\frac{1}{-n^2}\right)}\\ & =& lim\frac{-3\left(1+\frac{1}{3n}-\frac{2}{3n^2}\right)}{1+\frac{3}{n}-\frac{1}{n^2}}\\ & =& \frac{-3\left(1+\frac{1}{+\infty }-\frac{2}{+\infty }\right)}{1+\frac{3}{+\infty }-\frac{1}{+\infty }}\\ & =& \frac{-3\left(1+0-0\right)}{1+0-0}\\ & =& -3,\end{array}$$

$$\begin{array}{lll}lim{w}_n& =& lim\frac{3-2n}{n^2+n+1}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{-2n\left(\frac{3}{-2n}+1\right)}{n^2\left(1+\frac{n}{n^2}+\frac{1}{n^2}\right)}\\ & =& lim\frac{-2\left(-\frac{3}{2n}+1\right)}{n\left(1+\frac{1}{n}+\frac{1}{n^2}\right)}\\ & =& \frac{-2\left(-\frac{3}{+\infty }+1\right)}{+\infty \left(1+\frac{1}{+\infty }+\frac{1}{+\infty }\right)}\\ & =& \frac{-2\left(-0+1\right)}{+\infty \left(1+0+0\right)}\\ & =& \frac{-2}{+\infty }\\ & =& 0.\end{array}$$

The three different cases presented in Example 2.30 allow us to straightforwardly establish the following more general result.

Proposition 2.7. Let p(x) and q(x) be two non-null polynomials with degrees k and l and leading coefficients a and b, respectively. Let $\left(u_n\right)$ be the sequence defined by

$$u_n=\frac{p\left(n\right)}{q\left(n\right)},\forall n\in \mathbb{N}.$$

Then, the following statements hold.

1. If $k>l$ and $\frac{a}{b}>0$, then $lim{u}_n=+\infty ;$

2. If $k>l$ and $\frac{a}{b}<0$, then $lim{u}_n=-\infty ;$

3. If $k=l$, then $lim{u}_n=\frac{a}{b};$

4. If $k<l$, then $lim{u}_n=0$.

Next, we present two examples where an indeterminate form $\frac{\infty }{\infty }$ appears when evaluating the limit of sequences defined with radicals.

Example 2.31. Let $\left(u_n\right)$ and $\left(v_n\right)$ be the sequences defined by:

$$u_n=\frac{\sqrt{n}-2}{2\sqrt{n}+1}\mathit{\text{and}}{v}_n=\frac{2n+1}{\sqrt{n+2}}.$$

When evaluating the limit either of $u_n$ or $v_n$ by direct substitution we come upon the indeterminate form $\frac{\infty }{\infty }$. In each case, we will solve the indeterminate form by collecting terms and rewriting the fraction. Indeed, we have

$$\begin{array}{lll}lim{u}_n& =& lim\frac{\sqrt{n}-2}{2\sqrt{n}+1}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{\sqrt{n}\left(1-\frac{2}{\sqrt{n}}\right)}{\sqrt{n}\left(2+\frac{1}{\sqrt{n}}\right)}\\ & =& lim\frac{1-\frac{2}{\sqrt{n}}}{2+\frac{1}{\sqrt{n}}}\\ & =& \frac{1-\frac{2}{+\infty }}{2+\frac{1}{+\infty }}\\ & =& \frac{1-0}{2+0}\\ & =& \frac{1}{2},\end{array}$$

as well as

$$\begin{array}{lll}lim{v}_n& =& lim\frac{2n+1}{\sqrt{n+2}}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{2n\left(1+\frac{1}{2n}\right)}{\sqrt{n^2\left(\frac{n}{n^2}+\frac{2}{n^2}\right)}}\\ & =& lim\frac{2n\left(1+\frac{1}{2n}\right)}{n\sqrt{\frac{1}{n}+\frac{2}{n^2}}}\\ & =& lim\frac{2\left(1+\frac{1}{2n}\right)}{\sqrt{\frac{1}{n}+\frac{2}{n^2}}}\\ & =& \frac{2\left(1+\frac{1}{+\infty }\right)}{\sqrt{\frac{1}{+\infty }+\frac{2}{+\infty }}}\\ & =& \frac{2\left(1+0\right)}{\sqrt{0+0}}\\ & =& \frac{2}{0^{+}}\\ & =& +\infty .\end{array}$$

Note that the square root will always attain 0 through values bigger than 0 as it is not defined in ${\mathbb{R}}^{-}$.

In the following, we present an example of an indeterminate form $\frac{\infty }{\infty }$ that arises from dividing sequences of the form presented in Proposition 2.6.

Example 2.32. Consider the sequence $\left(u_n\right)$ with general term given by

$$u_n=\frac{4^n}{2^n+3^n}.$$

Clearly, evaluating the limit of $\left(u_n\right)$ by simple substitution results in the indeterminate form $\frac{\infty }{\infty }.$ In a similar manner than that of Example 2.28, we will solve this problem by collecting the higher power of the numerator and denominator like it follows.

$$\begin{array}{lll}lim{u}_n& =& lim\frac{4^n}{2^n+3^n}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{4^n}{3^n\left(\frac{2^n}{3^n}+1\right)}\\ & =& lim\left[\frac{4^n}{3^n}\times \frac{1}{{\left(\frac{2}{3}\right)}^n+1}\right]\\ & =& lim\left[{\left(\frac{4}{3}\right)}^n\times \frac{1}{{\left(\frac{2}{3}\right)}^n+1}\right]\\ & =& +\infty \left(\frac{1}{0+1}\right)\\ & =& +\infty \end{array}$$

Example 2.33. Let $\left(v_n\right)$ be the sequence defined by

$$v_n=\frac{n}{7^n}.$$

Attempting to evaluate the limit of $\left(v_n\right)$ directly gives us the indeterminate form $\frac{\infty }{\infty }$. Applying L’Hôpital’s Rule, to the function f, such that

$$f\left(x\right)=\frac{x}{7^x},$$

allows us to obtain the limit of $\left(v_n\right).$ Indeed,

$$\begin{array}{lll}\underset{x\to +\infty }{lim}f\left(x\right)& =& \underset{x\to +\infty }{lim}\frac{x}{7^x}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& \underset{x\to +\infty }{lim}\frac{\frac{d}{\mathit{dx}}\left(x\right)}{\frac{d}{\mathit{dx}}\left(7^x\right)}\\ & =& \underset{x\to +\infty }{lim}\frac{1}{7^{x}ln7}\\ & =& \frac{1}{+\infty }\\ & =& 0.\end{array}$$

Hence, by Theorem 2.10, $lim{v}_n=0$.

2.5.3 Indeterminate form $\frac{0}{0}$

While working out the limits of sequences, the indeterminate form $\frac{0}{0}$ arises mainly when dividing two sequences that converge to 0 and can be solved by simplifying the expression at hand, eventually obtaining a $\frac{\infty }{\infty }$ indeterminate form and applying some technique surveyed in the previous subsection.

Example 2.34. Consider the sequences $\left(u_n\right)$ and $\left(v_n\right)$ with general terms given by

$$u_n=\frac{1}{n+1}\mathit{\text{and}}{v}_n=\frac{3}{n^2-2}.$$

We have that

$$\begin{array}{lll}lim\frac{u_n}{v_n}& =& lim\frac{\frac{1}{n+1}}{\frac{3}{n^2-2}}\\ & \stackrel{\left(\frac{0}{0}\right)}{=}& lim\frac{n^2-2}{3\left(n+1\right)}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{n^2\left(1-\frac{2}{n^2}\right)}{3n\left(1+\frac{1}{n}\right)}\\ & =& lim\frac{n\left(1-\frac{2}{n^2}\right)}{3\left(1+\frac{1}{n}\right)}\\ & =& \frac{+\infty \left(1-\frac{2}{+\infty }\right)}{3\left(1+\frac{1}{+\infty }\right)}\\ & =& \frac{+\infty \left(1-0\right)}{3\left(1+0\right)}\\ & =& +\infty .\end{array}$$

Example 2.35. Let $\left(w_n\right)$ be the sequence defined by

$$w_n=\frac{ln\left(1-\frac{1}{n^2}\right)}{ln\left(1-\frac{2}{n^2}\right)}.$$

If we try to evaluate the limit of $\left(w_n\right)$ by direct substitution, we obtain the indeterminate form $\frac{0}{0}$.

Consider the function f such that

$$f\left(x\right)=\frac{ln\left(1-\frac{1}{x^2}\right)}{ln\left(1-\frac{2}{x^2}\right)}.$$

Applying L’Hôpital’s Rule, we have

$$\begin{array}{lll}\underset{x\to +\infty }{lim}f\left(x\right)& =& \underset{x\to +\infty }{lim}\frac{ln\left(1-\frac{1}{x^2}\right)}{ln\left(1-\frac{2}{x^2}\right)}\\ & \stackrel{\left(\frac{0}{0}\right)}{=}& \underset{x\to +\infty }{lim}\frac{\frac{d}{\mathit{dx}}\left[ln\left(1-\frac{1}{x^2}\right)\right]}{\frac{d}{\mathit{dx}}\left[ln\left(1-\frac{2}{x^2}\right)\right]}\\ & =& \underset{x\to +\infty }{lim}\frac{\frac{\frac{2}{x^3}}{1-\frac{1}{x^2}}}{\frac{\frac{4}{x^3}}{1-\frac{2}{x^2}}}\\ & =& \underset{x\to +\infty }{lim}\frac{1-\frac{2}{x^2}}{2\left(1-\frac{1}{x^2}\right)}\\ & =& \frac{1-0}{2\left(1-0\right)}\\ & =& \frac{1}{2}.\end{array}$$

Hence, by Theorem 2.10, $lim{w}_n=\frac{1}{2}$.

2.5.4 Indeterminate form $0\times \infty$

The case of the indeterminate form is solved using the techniques presented in the two previous subsections as it can always be converted into either the indeterminate form $\frac{\infty }{\infty }$ or $\frac{0}{0}$.

Example 2.36. Let $\left(u_n\right),\left(v_n\right)$ and $\left(w_n\right)$ be the sequences defined by

$$u_n=\frac{3}{n},v_n={\left(\frac{1}{2}\right)}^n\mathit{\text{and}}{w}_n=n-2.$$

We have:

$$\begin{array}{lll}lim\left(u_n\times w_n\right)& =& lim\frac{3}{n}\times \left(n-2\right)\\ & \stackrel{\left(0\times \infty \right)}{=}& lim\frac{3n-6}{n}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\frac{3n\left(1-\frac{6}{3n}\right)}{n}\\ & =& lim\left[3\left(1-\frac{2}{n}\right)\right]\\ & =& 3\left(1-\frac{2}{+\infty }\right)\\ & =& 3\left(1-0\right)\\ & =& 3.\end{array}$$

Also,

$$\begin{array}{lll}lim\left(v_n\times w_n\right)& =& lim{\left(\frac{1}{2}\right)}^n\times \left(n-2\right)\\ & \stackrel{\left(0\times \infty \right)}{=}& lim\frac{n-2}{2^n}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& lim\left[\frac{n}{2^n}-\frac{2}{2^n}\right]\\ & =& lim\frac{n}{2^n}-lim{2}^{1-n}\\ & =& lim\frac{n}{2^n}-0\\ & =& lim\frac{n}{2^n}.\end{array}$$E16

Now consider the real function f defined by

$$f\left(x\right)=\frac{x}{2^x},$$

applying L’Hôpital’s Rule, we have

$$\begin{array}{lll}\underset{x\to +\infty }{lim}f\left(x\right)& =& \underset{x\to +\infty }{lim}\frac{x}{2^x}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& \underset{x\to +\infty }{lim}\frac{1}{2^{x}ln2}\\ & =& \underset{x\to +\infty }{lim}\frac{1}{+\infty }\\ & =& 0.\end{array}$$

Therefore, we can conclude, by Theorem 2.10, that (16) is equal to 0.

Example 2.37. Let $\left(u_n\right)$ and $\left(v_n\right)$ be the sequences defined by

$$u_n=ln\left(n\right)\mathit{\text{and}}{v}_n=\frac{1}{n}.$$

We have that

$$\begin{array}{lll}lim\left(u_n\times v_n\right)& =& lim\left(ln\left(n\right)\times 1/n\right)\\ & \stackrel{\left(0\times \infty \right)}{=}& lim\frac{ln\left(n\right)}{n}.\end{array}$$E17

The limit (17) results in the indeterminate form $\frac{\infty }{\infty }$. As we did in previous examples, let us consider the function f such that

$$f\left(x\right)=\frac{ln\left(x\right)}{x}.$$

Applying L’Hôpital’s Rule, we obtain

$$\begin{array}{lll}\underset{x\to +\infty }{lim}f\left(x\right)& =& \underset{x\to +\infty }{lim}\frac{ln\left(x\right)}{x}\\ & \stackrel{\left(\frac{\infty }{\infty }\right)}{=}& \underset{x\to +\infty }{lim}\frac{\frac{d}{\mathit{dx}}\left(ln\left(x\right)\right)}{\frac{d}{\mathit{dx}}\left(x\right)}\\ & =& \underset{x\to +\infty }{lim}\frac{\frac{1}{x}}{1}\\ & =& \underset{x\to +\infty }{lim}\frac{1}{x}\\ & =& \frac{1}{+\infty }\\ & =& 0.\end{array}$$

Then, by Theorem 2.10, we can conclude that $lim\left(u_n\times v_n\right)=0.$

2.5.5 Indeterminate forms $0^0\mathbf{,}{\infty }^0$ and $1^{\infty }$

In this subsection, we present some examples for each of the remaining indeterminate forms, $0^0,{\infty }^0$_, and $1^{\infty }$, which were not addressed before. We apply the natural logarithm to the expressions to solve the indeterminate forms.

Example 2.38. In this example, we will calculate the value of

$$lim{\left(\frac{1}{n}\right)}^{\frac{1}{n}}.$$

We have:

$$\begin{array}{lll}lim{\left(\frac{1}{n}\right)}^{\frac{1}{n}}& \stackrel{\left(0^0\right)}{=}& lim\left[e^{ln{\left(\frac{1}{n}\right)}^{\frac{1}{n}}}\right]\\ & =& lim\left[e^{\frac{1}{n}ln\left(\frac{1}{n}\right)}\right]\\ & =& lim\left[e^{\frac{1}{n}\left(-ln\left(n\right)\right)}\right]\\ & =& e^{-lim\left[\frac{ln\left(n\right)}{n}\right]}\\ & =& e^0=1\end{array}$$

using Example 2.37.

Example 2.39. In this example, we will calculate the value of

$$lim{n}^{\frac{1}{n}}.$$

We have:

$$\begin{array}{lll}lim{n}^{\frac{1}{n}}& \stackrel{\left({\infty }^0\right)}{=}& lim\left[e^{ln\left(n^{\frac{1}{n}}\right)}\right]\\ & =& lim\left[e^{\frac{1}{n}ln\left(n\right)}\right]\\ & =& lim\left[e^{\frac{ln\left(n\right)}{n}}\right]\\ & =& e^{lim\left[\frac{ln\left(n\right)}{n}\right]}\\ & =& e^0=1\end{array}$$

using Example 2.37.

Example 2.40. In this example, we will calculate the value of

$$lim{\left(1+\frac{1}{n^2}\right)}^n.$$

We have:

$$\begin{array}{lll}lim{\left(1+\frac{1}{n^2}\right)}^n& \stackrel{\left(1^{\infty }\right)}{=}& lim\left[e^{ln{\left(1+\frac{1}{n^2}\right)}^n}\right]\\ & =& lim\left[e^{nln\left(1+\frac{1}{n^2}\right)}\right].\end{array}$$

Considering the real function f, defined by

$$f\left(x\right)=xln\left(1+\frac{1}{x^2}\right),$$

we have

$$\begin{array}{lll}\underset{x\to +\infty }{lim}f\left(x\right)& =& \underset{x\to +\infty }{lim}\left[xln\left(1+\frac{1}{x^2}\right)\right]\\ & \stackrel{\left(\infty \times 0\right)}{=}& \underset{y\to 0}{lim}\left[\frac{1}{y}ln\left(1+y^2\right)\right]\\ & \stackrel{\left(\frac{0}{0}\right)}{=}& \underset{y\to 0}{lim}\frac{\frac{d}{\mathit{dy}}\left(ln\left(1+y^2\right)\right)}{\frac{d}{\mathit{dy}}\left(y\right)}\\ & =& \underset{y\to 0}{lim}\frac{\frac{2y}{1+y^2}}{1}\\ & =& 0,\end{array}$$

changing variable $y=1/x$ and applying l’Hôpital’s Rule.

Hence, we conclude that the limit at hand is equal to $e^0=1$.

We finish this subsection with two results containing two notable cases for solving indeterminations treated here, $1^{\infty }$ and ${\infty }^0$.

Proposition 2.8. Let $\left(u_n\right)$ be a sequence of real numbers such that $u_n\ne 0$ for $n\ge m$, with $m\in \mathbb{N},$ converging to $\pm \infty$ and $a\in \mathbb{R}$. Then,

$$lim{\left(1+\frac{a}{u_n}\right)}^{u_n}=e^a.$$

Proposition 2.9. Let $a\in {\mathbb{R}}^{+}$ and $n\in \mathbb{N}$. Then,

1. $lim\sqrt[n]{a}=1;$

2. $lim\sqrt[n]{n}=1$.

2.6 Suggested exercises

General properties of sequences

1. Let $\left(u_n\right)$ be the sequence of real numbers 0, 2, 6, 12, 20, 30, …

   1. Find the general term of $\left(u_n\right).$

   2. Verify that 156 is a term of $\left(u_n\right).$ Find its order.

   3. Solve the condition $400<u_n<600$.

2. Let $\left(v_n\right)$ be the sequence whose general term is given by:

   $$v_n=\frac{6n+8}{3n+2}.$$

   1. Find the first five terms of $v_n$.

   2. Verify if $\frac{9}{4}$ is a term of $v_n$.

   3. Prove that $\left(v_n\right)$ is a decreasing sequence.

   4. Show that $\left(v_n\right)$ is bounded. Determine a lower bound and an upper bound of $\left(v_n\right)$.

   5. Determine the order from which $v_n$ is not greater than 2,1.

3. Consider the sequence $\left(w_n\right)$ whose first five terms are:

   $$-\frac{1}{4};\frac{8}{9};-\frac{27}{16};\frac{64}{25};-\frac{125}{36}$$

   1. Find the general term of $\left(w_n\right).$

   2. Determine $w_{10}$.

   3. Show that $\left(w_n\right)$ is unbounded.

4. Let $\left(u_n\right)$ be the sequence defined by recurrence as

   $$\left\{\begin{array}{l}{u}_1=5\\ u_{n+1}=3u_n+10,\forall n\in \mathbb{N}\end{array}\right..$$

   1. Find the first four terms of $\left(u_n\right).$

   2. Using mathematical induction, show that

      $$u_n=\frac{5}{3}\left(2\times 3^n-3\right),\forall n\in \mathbb{N}.$$

   3. Show that $\left(u_n\right)$ is monotonic.

   4. Prove that $\left(u_n\right)$ is not bounded from above.

5. Let $\left(v_n\right)$ be the sequence defined by

   $$v_n=sin\left(\frac{\mathit{n\pi }}{4}\right).$$

   1. Find the image of $\left(v_n\right).$

   2. Justify that $\left(v_n\right)$ is bounded.

   3. Justify that $\left(v_n\right)$ is not monotonic.

6. Let $\left(w_n\right)$ be the sequence defined by

   $$w_n=\left\{\begin{array}{ll}n-2,& \text{if}n\le 4\\ \frac{5n-4}{n},& \text{if}n>4\end{array}\right..$$

   1. Study the monotony of $\left(w_n\right).$

   2. Show that $\left(w_n\right)$ is bounded.

7. Consider the sequence $\left(u_n\right)$ of general term

   $$u_n=\frac{2n-7}{n+1}.$$

   Determine the number of positive terms of $\left(u_n\right)$ that are smaller than $\frac{7}{5}$.

8. Let $\left(v_n\right)$ be the sequence whose general term is

   $$v_n=\frac{1-2n}{2n+3}.$$

   1. Obtain the term of order 5 of $\left(v_n\right).$

   2. Verify if $-\frac{25}{29}$ is a term of $\left(v_n\right)$

   3. Study the monotony of $\left(v_n\right).$

   4. Show that $\left(v_n\right)$ is bounded.

9. Study the monotony of the sequence $\left(w_n\right)$ defined by $w_n={\left(-1\right)}^n+3n$.

   Arithmetic an geometric progressions

10. Let $\left(u_n\right)$ be the arithmetic sequence defined by:

    $$\left\{\begin{array}{l}{u}_1=2\\ u_{n+1}=u_n+3,\forall n\in \mathbb{N}\end{array}\right..$$

    1. Obtain the general term of $\left(u_n\right).$

    2. Determine the value of $u_{31}-u_{10}$.

    3. Determine the sum of the first 20 terms of the $\left(u_n\right).$

11. Let $\left(v_n\right)$ be the sequence whose general term is

    $$v_n=\frac{1-5n}{3}.$$

    1. Show that $\left(v_n\right)$ is an arithmetic progression.

    2. Justify that $\left(v_n\right)$ is a decreasing sequence.

    3. Determine the sum of the 15 consecutive terms of $\left(v_n\right)$ after $v_{22}$.

12. Let $\left(w_n\right)$ be an arithmetic progression, such that $w_5=12$ and $w_{13}=44$.

    1. Determine the common difference of $\left(w_n\right)$ and classify its monotony.

    2. Obtain the general term of $\left(w_n\right).$

    3. The sum of the first m terms of $\left(w_n\right)$ is 360. Determine the value of m.

13. Consider the geometric progression $\left(u_n\right)$ defined by

    $$\left\{\begin{array}{l}{u}_1=4\\ u_{n+1}=\frac{u_n}{5},\forall n\in \mathbb{N}\end{array}\right..$$

    1. Obtain the general term of $\left(u_n\right).$

    2. Determine the monotony of $\left(u_n\right).$

    3. Prove that $\left(u_n\right)$ is bounded.

    4. Compute the exact value of the sum of the first 10 terms of $\left(u_n\right).$

14. Consider the sequence $\left(v_n\right)$ whose general term is

    $$v_n=\frac{5}{2^{n-1}}.$$

    1. Show that $\left(v_n\right)$ is a geometric progression.

    2. Determine the monotony of $\left(v_n\right).$

    3. Show that $\left(v_n\right)$ is bounded.

    4. Obtain the exact value of $v_3+v_4+\cdots +v_9$.

15. Consider the arithmetic progression $\left(w_n\right)$ defined by

    $$\left\{\begin{array}{l}{w}_1=-50\\ w_{n+1}=w_n+2,\forall n\in \mathbb{N}\end{array}\right..$$

    For a certain $p\in \mathbb{N}$, the sum of the first p terms of $\left(w_n\right)$ is equal to p. Determine the value of p.

    Limits and convergence

16. Let $\left(u_n\right)$ be the sequence defined by

    $$u_n=\frac{2n}{n+3}.$$

    Using the definition of convergent sequence, show that $lim{u}_n=2$.

17. Consider the sequence $\left(v_n\right)$ such that

    $$v_n=\frac{2cos\left(\mathit{n\pi }\right)}{\sqrt{n+1}}.$$

    Using the Squeeze Theorem for Convergent Sequences, show that

    $$lim{v}_n=0.$$

18. Evaluate $lim{w}_n$ where

    1. $w_n=-4n^2+n-5$

    2. $w_n=7n^5-100n^2+500$

    3. $w_n=\frac{3n^5-2n^2+3n-7}{2n^3-n^2+2n-3}$

    4. $w_n=\frac{3n^3+n^2-5}{2n^3-3n^2-n+6}$

    5. $w_n=\frac{n^2+3n-5}{-2n^3-5n^2+n+2}$

    6. $w_n=\frac{\sqrt{2n+5}-\sqrt{n}}{n+5}$

    7. $w_n=\frac{2n}{\sqrt{n^2+1}+3n}$

    8. $w_n=\frac{\sqrt{2}{n}^3-\sqrt{3}n}{-n^4+4n-3}$

    9. $w_n=\sqrt{n^2-1}-n+3$

    10. $w_n=\frac{ln\left(2n+5\right)}{n^2-1}$

    11. $w_n=\frac{sin\left(n-2\right)}{n+1}$

    12. $w_n=7^n-2^n$

19. Consider the sequences $\left(u_n\right)$ and $\left(v_n\right)$ whose general terms are given by

    $$u_n={\left(-1\right)}^n\text{and}{v}_n=\frac{3}{2n}.$$

    Calculate limit of the sequence $\left(u_n\times v_n\right).$

20. Let $\left(w_n\right)$ be a monotonic geometric progression such that $w_5=125$ and $w_{11}=\frac{1}{125}$.

    1. Show that $w_1=78125$ and $r=\frac{1}{5}$, where r is the common ratio of $\left(w_n\right).$

    2. Let $S_n$ be the sum of the first n terms of $\left(w_n\right).$ Calculate $lim{S}_n$.

21. Consider the sequence $\left(u_n\right)$ whose general term is given by

    $$u_n=\frac{3}{\left(n+1\right)!5^n}.$$

    Show that $\left(u_n\right)$ is a Cauchy sequence.

    Miscellaneous problems

22. Determine the monotony of the sequence $\left(v_n\right)$ as defined below.

    1. $v_n=\frac{1}{2n+1};$

    2. $v_n=\frac{n+2}{n+1};$

    3. $v_n=\frac{n}{e^n};$

    4. $v_n=\frac{n}{n^2+1}.$

23. Consider the sequence $\left(w_n\right)$ such that

    $$w_n=\frac{n!}{n^n}$$

    for all $n\in \mathbb{N}$.

    1. Show that $\left(w_n\right)$ is decreasing.

    2. Justify that $\left(w_n\right)$ is convergent.

24. Using the definition of limit, prove the following equalities.

    1. $lim\frac{n+1}{2n+1}=\frac{1}{2};$

    2. $lim\frac{n^4-1}{n^4+1}=1;$

    3. $lim\frac{sin\left(n\right)}{n^2}=0;$

    4. $lim\frac{ln\left(n\right)}{n}=0;$

    5. $lim{\left(\frac{1+n}{n}\right)}^n=e.$

25. Let $\left(u_n\right)$ be a sequence of real numbers converging to 1. Consider the new sequence $\left(v_n\right)$ such that

    $$v_n=\frac{{\left(u_n\right)}^3-1}{u_n-1}.$$

    Analyze the convergence of $\left(v_n\right).$

26. Show that

    $$lim\frac{1^2+2^2+\cdots +n^2}{n^4}=0.$$

27. Calculate the following limits:

    1. $lim{\left(\frac{n+1}{n+2}\right)}^n;$

    2. $lim\left(\sqrt{n+1}-\sqrt{n}\right);$

    3. $lim\sqrt[n]{n};$

    4. $lim\frac{2^n}{n};$

    5. $lim\left(\frac{1}{\sqrt[3]{n}}+\frac{1}{\sqrt[3]{n+1}}+\cdots +\frac{1}{\sqrt[3]{2n}}\right);$

    6. $lim\sqrt[n]{3};$

    7. $lim\frac{2^n+3^n}{4^n+7^n};$

    8. $lim\frac{7^n+8^n}{3^n+4^n};$

    9. $lim\sqrt[n]{3^n+9^n}.$

28. Show that, if $\left(w_n\right)$ is a sequence of positive real numbers such that $lim\frac{w_{n+1}}{w_n}=L\in \mathbb{R},$ then $lim\sqrt[n]{w_n}=L.$

29. Prove that, given $\left(u_n\right)$ a convergent sequence of real numbers, if $lim{u}_n=L$, then

    $$lim\frac{u_1+u_2+\cdots +u_n}{n}=L.$$

30. Consider the sequence $\left(v_n\right)$ such that $v_1=\sqrt{2}$ and $v_{n+1}=\sqrt{2+v_n}$ for $n\ge 2$.

    1. Show that $\left(v_n\right)$ is an increasing sequence.

    2. Show that $\left(v_n\right)$ is bounded from above.

    3. Show that $lim{v}_n=2$.

31. Show that the sequence $\left(w_n\right)$ of positive terms, where $w_1=\sqrt{2}$ and $w_{n+1}=\sqrt{2w_n}$ for $n\in \mathbb{N}\backslash \left\{1\right\}$, is convergent. Find its limit.

32. Show that

    $$lim\left(\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\cdots +\frac{1}{\sqrt{2n+1}}\right)=+\infty .$$

33. Show that the sequence $\left(u_n\right),$ where

    $$u_n=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+n}},$$

    for all $n\in \mathbb{N}$, is convergent.

34. Show that

    $$lim\left(\sqrt[3]{n+1}-\sqrt[3]{n}\right)=0.$$

35. Use mathematical induction to show that, $\forall n\in \mathbb{N}$, we have:

    1. $1+3+5+\cdots +2n-1=n^2;$

    2. $\sum _{k=1}^n{k}^3={\left(\sum _{k=1}^{n}k\right)}^2;$

    3. $\sum _{k=1}^n\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}.$

36. Consider a convergent sequence $\left(u_n\right)$ such that $u_n\ne 2$, for all $n\in \mathbb{N}$ and $lim{u}_n=2.$ Calculate

    $$lim\frac{u_n+u_n^2+u_n^3-14}{u_n-2}.$$

37. Show that the sequences $\left(v_n\right)$ are decreasing, where:

    1. $v_n=\frac{ln\left(n+2\right)}{n+2};$

    2. $v_n=\frac{n}{9^n}.$

3. Series

This section is devoted to the subject of series, which are essentially generalized infinite sums of numbers. In view of this, it is easy to understand the relevance of having addressed the subject of sequences in the previous section. Indeed, we can say that the following survey on series will be made on the shoulders of all that was presented about sequences of real numbers.

The subject of series plays a major role in mathematical analysis, and it is widely applied to many different areas like economics and finance, statistics, physics, or computer science.

This section is organized as follows. We start by presenting, on Section 3.1, the basic definitions and results involving series, including convergence. Then, from Sections 3.2 to 3.5 we present some important types of series, including series of non-negative terms, alternating series, function and power series, along with the application to the representation of functions. As in the previous section, we propose, in Section 3.6, a set of exercises for the reader to grasp their acquired knowledge on this subject, with which we finish this section.

3.1 Basic definitions and convergence

We start this section by defining the concept of series by means of the concept of sequence. Before, however, note that along this text, we will only consider series of real numbers.

Definition 3.1 (Series). Given a sequence of real numbers, $\left(u_n\right)$, a series is an expression of the form

$$\sum _{k=1}^{+\infty }{u}_k=u_1+u_2+\cdots ,$$

where each $u_k,k\in \mathbb{N}$, is simultaneous a term of the series and a term of the sequence $\left(u_n\right)$.

The sequence $\left(u_n\right),$ of Definition 3.1, is sometimes called the generating sequence of the series, due to its nature.

One can also come to an equivalent definition of series if, from $\left(u_n\right),$ we build a new sequence $\left(s_n\right),$ with general term

$$s_n=u_1+u_2+\cdots +u_n,$$

which is called the sequence of partial sums of $\left(u_n\right).$ Note that the value of the series ${\sum }_{k=1}^{+\infty }{u}_k$ of Definition 3.1, also referred to as the sum of the series, is the same as calculating ${lim}_{n\to +\infty }{s}_n$. The next definition is, therefore, natural.

Definition 3.2 (Convergent series). Let $\left(u_n\right)$ be a sequence and $\left(s_n\right)$ be its corresponding partial sums sequence. Then, we say that the series ${\sum }_{k=1}^{+\infty }{u}_k$ is convergent (divergent) if and only if the sequence $\left(s_n\right)$ is convergent (divergent).

Next, we present some examples where Definition 3.2 is put to effective use.

Example 3.1. The series ${\sum }_{k=1}^{+\infty }1$ is divergent. Indeed, it is only necessary to observe that the sequence of partial sums of the constant sequence with all terms equal to 1 is given by the general term $s_n=n$ and, for this sequence, we have that $lim{s}_n=limn=+\infty$. Since $\left(s_n\right)$ is divergent, then the corresponding series ${\sum }_{k=1}^{+\infty }1$ is divergent.

Example 3.2. Let’s study the convergence of the series

$$\sum _{n=1}^{+\infty }{\left(\frac{1}{2}\right)}^n.$$

The generating sequence of this series is a geometric sequence with the first term and common ratio both equal to $\frac{1}{2}$. Then, by applying the sum formula for a geometric sequence, the partial sums sequence is given by

$$s_n=\frac{1}{2}\left(\frac{1-{\left(\frac{1}{2}\right)}^n}{1-\frac{1}{2}}\right).$$

Then, we have

$$\begin{array}{lll}lim{s}_n& =& \frac{1}{2}\times lim\left(\frac{1-{\left(\frac{1}{2}\right)}^n}{1-\frac{1}{2}}\right)\\ & =& lim\left(1-{\left(\frac{1}{2}\right)}^n\right)\\ & =& 1.\end{array}$$

Therefore, $\left(s_n\right)$ is convergent and, consequently, the series ${\sum }_{n=1}^{+\infty }{\left(\frac{1}{2}\right)}^n$ is also convergent and we write

$$\sum _{n=1}^{+\infty }{\left(\frac{1}{2}\right)}^n=1.$$

Example 3.3. Consider the series ${\sum }_{n=1}^{+\infty }{5}^n$.

Like in Example 3.2, this series also has a generating sequence that is a geometric progression. In this case, the first term and the common ratio are both equal to 5. Thus, we have

$$\begin{array}{lll}lim{s}_n& =& lim5\left(\frac{1-5^n}{1-5}\right)\\ & =& -\frac{5}{4}lim\left(1-5^n\right)\\ & =& -\frac{5}{4}\left[1-\left(+\infty \right)\right]\\ & =& -\frac{5}{4}\left(-\infty \right).\\ & =& +\infty .\end{array}$$

Since $\left(s_n\right)$ is divergent, the series ${\sum }_{n=1}^{+\infty }{5}^n$ is also divergent. Some authors write ${\sum }_{n=1}^{+\infty }{5}^n=+\infty ,$ though.

Next, we present an example of a series whose generating sequence is an alternating sequence.

Example 3.4. Consider the series ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^{n+1}$.

For the corresponding sequence, the partial sums are:

$$\begin{array}{lll}{s}_1& =& 1\\ s_2& =& 1-1=0\\ s_3& =& 1-1+1=1\\ s_4& =& 1-1+1-1=0\\ s_5& =& 1-1+1-1+1=1\\ & ⋮& \\ s_{2n-1}& =& 1\\ s_{2n}& =& 1.\end{array}$$

Since $lim{s}_{2n-1}=1$ and $lim{s}_{2n}=0,lim{s}_n$ does not exist and, therefore, the series ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^{n+1}$ is divergent.

Example 3.5. Let’s analyze the convergence of the series of the form

$$\sum _{n=1}^{+\infty }{r}^{n-1},$$

with $r\in \mathbb{R}$, also known as a geometric series.

The corresponding sequence of partial sums, $\left(s_n\right)$ has its general term given by

$$s_n=\sum _{k=1}^n{r}^{k-1}.$$

We will break down this analysis into different situations.

1. Let $r\ne 1$. Then, we have the following equalities:

   $$s_n=1+r+r^2+\cdots +r^{n-1}$$E18

   and

   $${\mathit{rs}}_n=r+r^2+\cdots +r^n,$$E19

   Subtracting (19) from (18) yields

   $$\begin{array}{ll}& s_n\left(1-r\right)=1-r^n\\ \iff & s_n=\frac{1-r^n}{1-r}.\end{array}$$

   Observe that the convergence of $\left(s_n\right),$ and, therefore, of the series, depends now on the fact that $\mid r\mid$ is smaller or bigger than 1.

   1. If $\mid r\mid <1$, we have that $lim{r}^n=0$. Then,

      $$\begin{array}{lll}lim{s}_n& =& lim\frac{1-r^n}{1-r}\\ & =& \frac{1}{1-r}.\end{array}$$

      Then the series is convergent and its sum is equal to $\frac{1}{1-r}$.

   2. If $\mid r\mid >1$, there are two cases two consider, $r>1$ and $r<-1$.

      1. If $r>1$, then $lim{r}^n=+\infty$ and we have

         $$\begin{array}{lll}lim{s}_n& =& lim\frac{1-r^n}{1-r}\\ & =& \frac{1-\left(+\infty \right)}{1-r}\\ & =& \frac{1-\infty }{1-r}\\ & =& +\infty ,\end{array}$$

         and we conclude that the series is divergent.

      2. If $r<-1$, then $lim{r}^n$ does not exist since $lim{r}^{2n-1}=-\infty$ and $lim{r}^{2n}=+\infty .$ Therefore,

         $$\begin{array}{lll}lim{s}_n& =& lim\frac{1-r^n}{1-r}\\ & =& \frac{1-lim{r}^n}{1-r}\end{array}$$

         Does not exist and the series is divergent.

   3. We are missing to consider one single sub-case from the general case $r\ne 1$, that is, the case when $r=-1$. If $r=-1$, then

      $$\begin{array}{lll}lim{s}_n& =& lim\frac{1-{\left(-1\right)}^n}{1-\left(-1\right)}\\ & =& lim\frac{1-{\left(-1\right)}^n}{2}.\end{array}$$

      Now, since

      $$lim{s}_{2n-1}=lim\frac{1-\left(-1\right)}{1-\left(-1\right)}=1$$

      and

      $$lim{s}_{2n}=lim\frac{1-1}{1+1}=0,$$

      we conclude that $lim{s}_n$ does not exist and the series is divergent.

2. The final case to consider is, of course, the case when $r=1$. If $r=1$, we have

$$\begin{array}{lll}lim{s}_n& =& lim\sum _{k=1}^n{1}^k\\ & =& limn\\ & =& +\infty .\end{array}$$

Thus, the series diverges.

Hence, we conclude that the series ${\sum }_{n=1}^{+\infty }{r}^{n-1}$ is convergent if and only if $\mid r\mid <1$ and, in this case, we have

$$\sum _{n=1}^{+\infty }{r}^{n-1}=\frac{1}{1-r}.$$

The general case of the geometric series is presented next.

Proposition 3.1. Let $a,r\in \mathbb{R}$. If $\mid r\mid <1$, then the geometric series of first term a and common ratio r,

$$\sum _{n=1}^{+\infty }{\mathit{ar}}^{n-1},$$

is convergent and its sum is $\frac{a}{1-r}$.

In some cases, the properties of the generating sequence allow us to deduce easily an expression for the partial sums sequence.

Example 3.6. Consider the series ${\sum }_{n=1}^{+\infty }\left(\sqrt{n}-\sqrt{n+1}\right)$.

Observe that

$$\begin{array}{lll}{s}_1& =& \sqrt{1}-\sqrt{2}\\ s_2& =& \left(\sqrt{1}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{3}\right)\\ s_3& =& \left(\sqrt{1}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{4}\right)\\ & ⋮& \\ s_n& =& \left(\sqrt{1}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{4}\right)\\ & & +\cdots +\left(\sqrt{n}-\sqrt{n+1}\right)\\ & =& 1-\sqrt{n+1}\end{array}$$

Hence, $lim{s}_n=-\infty$ and, therefore, the series ${\sum }_{n=1}^{+\infty }\left(\sqrt{n}-\sqrt{n+1}\right)$ is divergent.

Example 3.7. Consider the series

$$\sum _{n=1}^{+\infty }\frac{1}{n\left(n+1\right)}.$$

Firstly, note that

$$\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}.$$

Then, we have

$$\begin{array}{lll}{s}_1& =& \left(1-\frac{1}{2}\right)\\ s_2& =& \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)\\ s_3& =& \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)\\ & ⋮& \\ s_n& =& \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\\ & & +\cdots +\left(\frac{1}{n-1}-\frac{1}{n}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ & =& 1-\frac{1}{n+1}.\end{array}$$

Hence, $lim{s}_n=lim\left(1-\frac{1}{n+1}\right)=1$ and the series ${\sum }_{n=1}^{+\infty }\frac{1}{n\left(n+1\right)}$ is convergent with sum equal to 1.

We will now introduce a special kind of series.

Definition 3.3 (Mengoli’s series). A series ${\sum }_{n=1}^{+\infty }{u}_n$ such that $u_n=v_n-v_{n+m}$, with $m\in \mathbb{N}$, is called Mengoli’s series.

Mengoli’s series is also sometimes referred to as Telescopian series. The convergence of Mengoli’s series is easy to analyze.

Proposition 3.2. A Mengoli’s series ${\sum }_{n=1}^{+\infty }\left(v_n-v_{n+m}\right),m\in \mathbb{N}$, is convergent if and only if the sequence $\left(v_n\right)$ is convergent. Additionally,

$$\sum _{n=1}^{+\infty }\left(v_n-v_{n+m}\right)=\sum _{k=1}^m{v}_k-mlim{v}_n.$$

Proof. For Mengoli’s series ${\sum }_{n=1}^{+\infty }\left(v_n-v_{n+m}\right)$, we have that the partial sums sequence, $\left(s_n\right),$ is given by

$$\begin{array}{lll}{s}_n& =& \sum _{k=1}^n\left(v_k-v_{k+m}\right)\\ & =& \sum _{k=1}^n{v}_k-\sum _{k=1}^n{v}_{k+m}\\ & =& \sum _{k=1}^n{v}_k-\sum _{k=m+1}^{n+m}{v}_k\\ & =& \sum _{k=1}^m{v}_k+\sum _{k=m+1}^n{v}_k-\sum _{k=m+1}^n{v}_k-\sum _{k=n+1}^{n+m}{v}_k\\ & =& \sum _{k=1}^m{v}_k-\sum _{k=n+1}^{n+m}{v}_k.\end{array}$$

Now, let $\left(w_n\right)$ be the sequence whose general term is

$$w_n=v_{n+1}+v_{n+2}+\cdots +v_{n+m}.$$

Observe that the series ${\sum }_{n=1}^{+\infty }\left(v_n-v_{n+m}\right)$ converges if and only if $\left(w_n\right)$ converges and, in this case, we have

$$\begin{array}{lll}lim{w}_n& =& lim\left(v_{n+1}+v_{n+2}+\cdots +v_{n+m}\right)\\ & =& lim{v}_{n+1}+lim{v}_{n+2}+\cdots +lim{v}_{n+m}\\ & =& mlim{v}_n.\end{array}$$

Hence,

$$\begin{array}{lll}\sum _{n=1}^{+\infty }\left(v_n-v_{n+m}\right)& =& lim{s}_n\\ & =& lim\left(\sum _{k=1}^m{v}_k-\sum _{k=n+1}^{n+m}{v}_k\right)\\ & =& \sum _{k=1}^m{v}_k-lim\sum _{k=n+1}^{n+m}{v}_k\\ & =& \sum _{k=1}^m{v}_k-mlim{v}_n.\end{array}$$

□

The particular case of a Mengoli’s series with $m=1$ is worth considering.

Corollary 3.1. A Mengoli’s series ${\sum }_{n=1}^{+\infty }\left(v_n-v_{n+1}\right)$ is convergent if and only if the sequence $\left(v_n\right)$ is convergent. Additionally,

$$\sum _{n=1}^{+\infty }\left(v_n-v_{n+1}\right)=v_1-lim{v}_n.$$

We proceed with some relevant properties involving series convergence.

Theorem 3.1. Let $\alpha$ be a real non null number. Then ${\sum }_{n=1}^{+\infty }{u}_n$ converges if and only if ${\sum }_{n=1}^{+\infty }\left(\alpha u_n\right)$ converges.

Proof. Let $\left(s_n^1\right)$ and $\left(s_n^2\right)$ denote the sequences of partial sums of the series ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=1}^{+\infty }\left(\alpha u_n\right)$, respectively.

Since

$$\begin{array}{lll}{s}_n^2& =& \sum _{k=1}^n\left(\alpha u_k\right)\\ & =& \alpha \sum _{k=1}^n{u}_k\\ & =& \alpha s_n^1,\end{array}$$

then we have that

$$lim{s}_n^2=\alpha lim{s}_n^1$$

and, therefore,

$$\sum _{n=1}^{+\infty }\left(\alpha u_n\right)=\alpha \sum _{n=1}^{+\infty }{u}_n.$$

□

Theorem 3.2. Let $\alpha$ and $\beta$ be non null real numbers. If ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=1}^{+\infty }{v}_n$ are convergent series, then the series ${\sum }_{n=1}^{+\infty }\left(\alpha u_n+\beta v_n\right)$ is also a convergent series.

Proof. Let $\left(s_n^1\right)$ and $\left(s_n^2\right)$ denote the partial sums sequences of the series ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=1}^{+\infty }{v}_n,$ respectively. Also, since both series are convergent, let $lim{s}_n^1=S_1$ and $lim{s}_n^2=S_2$, for some $S_1,S_2\in \mathbb{R}$.

We have:

$$\begin{array}{lll}\sum _{k=1}^n\left(\alpha u_k+\beta v_k\right)& =& \sum _{k=1}^n\alpha u_k+\sum _{k=1}^n\beta v_k\\ & =& \alpha \sum _{k=1}^n{u}_k+\beta \sum _{k=1}^n{v}_k\\ =\alpha s_n^1+\beta s_n^2.& & \end{array}$$

Hence,

$$\begin{array}{lll}lim\left(\alpha s_n^1+\beta s_n^2\right)& =& \alpha lim{s}_n^1+\beta lim{s}_n^2\\ & =& \alpha S_1+\beta S_2.\end{array}$$

We conclude that the series ${\sum }_{n=1}^{+\infty }\left(\alpha u_n+\beta v_n\right)$ is convergent and that the sum of this series is $\alpha S_1+\beta S_2$ where $S_1$ and $S_2$ are the sum of the series ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=1}^{+\infty }{v}_n$, respectively. □

Note that if we consider two divergent series, ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=1}^{+\infty }{v}_n$, we cannot conclude that the series ${\sum }_{n=1}^{+\infty }\left(\alpha u_n+\beta v_n\right)$ is also divergent, for any real numbers $\alpha$ and $\beta$. Indeed, we have the series ${\sum }_{n=1}^{+\infty }\frac{1}{n}$ and ${\sum }_{n=1}^{+\infty }\left(-\frac{1}{n}\right)$ are both divergent and yet the series ${\sum }_{n=1}^{+\infty }\left(\frac{1}{n}-\frac{1}{n}\right)$ is clearly convergent to 0.

Theorem 3.3. Let l be a natural number. Then, the series ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=l+1}^{+\infty }{u}_n$ are either both convergent or both divergent.

Proof. Let $\left(s_n^1\right)$ and $\left(s_n^2\right)$ denote the partial sums sequences: $s_n^1={\sum }_{k=1}^n{u}_k$ and $s_n^2={\sum }_{k=l+1}^n{u}_k$ and let $S_1=lim{s}_n^1$ and $S_2=lim{s}_n^2$.

If n is a natural number greater than l, we have

$$\begin{array}{lll}{s}_n^1& =& \sum _{k=1}^n{u}_k\\ & =& \sum _{k=1}^l{u}_k+\sum _{k=l+1}^n{u}_k\\ & =& \sum _{k=1}^l{u}_k+s_n^2.\end{array}$$

Applying limits, we obtain

$$lim{s}_n^1=\sum _{k=1}^l{u}_k+lim{s}_n^2$$E20

From equality (20), we straightforwardly conclude that the series ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=l+1}^{+\infty }{u}_n$ are both convergent or both divergent, and, in the first case, we have that

$$S_1=\sum _{k=1}^l{a}_k+S_2.$$

□

It is a rather common situation to find ourselves with a series at hand for which there is not easy way to obtain an expression for the respective partial sums sequence. In these situations it is important to establish a criterium that allows us to identify convergent series without having to resort to the partial sums sequence. Such a criterium, that is, a necessary condition for the convergence of series, is presented next.

Theorem 3.4. If the series ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent, then $lim{u}_n=0$.

Proof. Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a convergent series, such that ${\sum }_{n=1}^{+\infty }{u}_n=l,l\in \mathbb{R}$, and let $\left(s_n\right)$ be its sequence of partial sums.

Since $u_n=s_n-s_{n-1}$ and $lim{s}_n=lim{s}_{n-1}=l$, we have

$$\begin{array}{lll}lim{u}_n& =& lim{s}_n-lim{s}_{n-1}\\ & =& l-l\\ & =& 0.\end{array}$$

□

Example 3.8. Let’s prove that the series

$$\sum _{n=1}^{+\infty }\frac{2^n}{2^n+1}$$

is divergent.

Let $\left(u_n\right)$ be the sequence whose general term is given by

$$u_n=\frac{2^n}{2^n+1}.$$

We have that

$$\begin{array}{lll}lim{u}_n& =& lim\frac{2^n}{2^n+1}\\ & =& lim\frac{2^n}{2^n\left(1+\frac{1}{2^n}\right)}\\ & =& lim\frac{1}{1+\frac{1}{2^n}}\\ & =& \frac{1}{1+0}\\ & =& 1\ne 0.\end{array}$$

Then, by Theorem 3.4, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{2^n}{2^n+1}$ is divergent.

3.2 Series of non-negative terms

In this section we focus on series ${\sum }_{n=1}^{+\infty }{u}_n$, such that $u_n\ge 0,\forall n\in \mathbb{N}$, which are called series of non-negative terms, and we survey the most relevant techniques for studying their convergence.

Theorem 3.5. Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a sequence of non-negative terms. Then ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent if and only if its sequence of partial sums is bounded from above.

Proof. Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of non-negative terms with $\left(s_n\right)$ being the respective partial sums sequence.

Let ${\sum }_{n=1}^{+\infty }{u}_n$ be convergent with $lim{s}_n=L.$ Note that $s_{n+1}=s_n+u_{n+1}$ and $u_{n+1}\ge 0$ by hypothesis. Then, clearly, $\left(s_n\right)$ is an increasing sequence and $s_n\le L$, that is, $\left(s_n\right)$ is bounded from above.

Conversely, let $\left(s_n\right)$ be bounded from above. Since $\left(s_n\right)$ is an increasing sequence bounded from above, then it is convergent to the supremum of its range, $L\in \mathbb{R}$. Then, $lim{s}_n=L$ and ${\sum }_{n=1}^{+\infty }{u}_n$ is a convergent series. □

Example 3.9. Let’s show that the Harmonic series, ${\sum }_{n=1}^{+\infty }\frac{1}{n}$, is divergent by proving that the corresponding partial sums sequence $\left(s_n\right)$ is divergent.

We have that

$$\begin{array}{lll}{s}_{2n}& =& \sum _{k=1}^{2n}\frac{1}{k}\\ & =& \sum _{k=1}^n\frac{1}{k}+\sum _{k=n+1}^{2n}\frac{1}{k}\\ & =& s_n+\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}\\ & \ge & s_n+\frac{1}{2n}+\frac{1}{2n}+\cdots +\frac{1}{2n}\\ & =& s_n+\frac{n}{2n}\\ & =& s_n+\frac{1}{2}\end{array}$$

and, therefore,

$$s_{2n}-s_n\ge \frac{1}{2}$$

Now, suppose that $lim{s}_n=lim{s}_{2n}=L$, with $L\in \mathbb{R}$. Then, we would have

$$lim{s}_{2n}-lim{s}_n=L-L=0\ge \frac{1}{2},$$

which is absurd. Therefore, we conclude that such an L does not exist and, accordingly, the sequence $\left(s_n\right)$ is divergent. Hence, the harmonic series is also divergent.

Theorem 3.6 (Comparison test). Let ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=1}^{+\infty }{v}_n$ be two series of non-negative terms such that $u_n\le v_n,\forall n\in \mathbb{N}$. Then we have:

1. If ${\sum }_{n=1}^{+\infty }{v}_n$ converges then ${\sum }_{n=1}^{+\infty }{u}_n$ converges.

2. If ${\sum }_{n=1}^{+\infty }{u}_n$ diverges then ${\sum }_{n=1}^{+\infty }{v}_n$ diverges.

Proof. Let ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=1}^{+\infty }{v}_n$ be two series of non-negative terms such that $u_n\le v_n,\forall n\in \mathbb{N}.$ Denote by $\left(s_n^1\right)$ and $\left(s_n^2\right)$ denote the partial sums sequences of the series ${\sum }_{n=1}^{+\infty }{u}_n$ and ${\sum }_{n=1}^{+\infty }{v}_n,$ respectively.

1. If ${\sum }_{n=1}^{+\infty }{v}_n$ is convergent, let $lim{s}_n^2=S$, with $S\in \mathbb{R}$. Then, we have the following inequalities:

   $$s_n^1\le s_n^2\le S.$$

   Since $\left(s_n^2\right)$ is an increasing sequence that is bounded from above, we have that $\left(s_n^1\right)$ is convergent and, consequently, the series ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent.

2. If ${\sum }_{n=1}^{+\infty }{u}_n$ is divergent, since it is a series of non-negative terms, then $lim{s}_n^1=+\infty .$ Therefore, $lim{s}_n^2=+\infty$ and the series ${\sum }_{n=1}^{+\infty }{v}_n$ is divergent.

□

Theorem 3.6 is usually referred to as the comparison test. Note that it is enough that there exists a $m\in \mathbb{N}$ such that $u_n\le v_n,\forall n\ge m$, for Theorem 3.6’s assertions to be valid.

We will now present a couple of examples where we apply Theorem 3.6.

Example 3.10. In this example, we will study the convergence of the series:

$$\sum _{n=1}^{+\infty }\frac{ln\left(n+2\right)}{n+2}.$$

Firstly, note that the following inequality holds for every natural n:

$$\frac{ln\left(n+2\right)}{n+2}\ge \frac{ln\left(3\right)}{n+2}.$$

Secondly, we have that

$$\sum _{n=1}^{+\infty }\frac{ln\left(3\right)}{n+2}=ln\left(3\right)\sum _{n=1}^{+\infty }\frac{1}{n+2}=ln\left(3\right)\sum _{n=3}^{+\infty }\frac{1}{n}.$$

Since ${\sum }_{n=3}^{+\infty }\frac{1}{n}$ is just the Harmonic series starting at the third term, which we already saw on Example 3.9, that it is not convergent, we can conclude that the series ${\sum }_{n=1}^{+\infty }\frac{ln\left(3\right)}{n+2}$ is also divergent.

Finally, using the comparison test, Theorem 3.6-(b), we can conclude that the series

$$\sum _{n=1}^{+\infty }\frac{ln\left(n+2\right)}{n+2}$$

is divergent.

Example 3.11. Let’s analyze the convergence of the series:

$$\sum _{n=1}^{+\infty }\left(\frac{3^n}{3^{2n}+1}\right).$$

Observe that

$$\begin{array}{lll}\frac{3^n}{3^{2n}+1}& \le & \frac{3^n}{3^{2n}}\\ & \le & \frac{3^n}{3^{2n}}\\ & =& \frac{3^n}{{\left(3^2\right)}^n}\\ & =& {\left(\frac{1}{3}\right)}^n.\end{array}$$

Now, since the series ${\sum }_{n=1}^{+\infty }{\left(\frac{1}{3}\right)}^n$ converges because it is a geometric series with ratio $\frac{1}{3}$ and absolute value smaller than 1, then, using the comparison test, Theorem 3.6-(a), we conclude that the series

$$\sum _{n=1}^{+\infty }\left(\frac{3^n}{3^{2n}+1}\right)$$

is also convergent.

Theorem 3.7 (Cauchy condensation test). Let $\left(u_n\right)$ be a decreasing sequence of non-negative terms. Then the series ${\sum }_{n=1}^{+\infty }{u}_n$ converges if and only if the series ${\sum }_{n=0}^{+\infty }{2}^n{u}_{2^n}$ converges.

Proof. Let $\left(u_n\right)$ be a decreasing sequence of non-negative terms.

1. Suppose that the series ${\sum }_{n=0}^{+\infty }{2}^n{u}_{2^n}$ converges.

   Since the sequence $\left(u_n\right)$ is decreasing, then the following inequalities hold:

   $$\begin{array}{lll}{u}_2+u_3& \le & u_2+u_2=2^1{u}_{2^1}\\ u_4+u_5+u_6+u_7& \le & u_4+u_4+u_4+u_4=2^2{u}_{2^2}\\ u_8+u_9+\cdots +u_{15}& \le & u_8+u_8+\cdots +u_8=2^3{u}_{2^3}\\ u_{16}+u_{17}+\cdots +u_{31}& \le & 2^{16}{u}_{2^{16}}\\ & ⋮& \\ u_{2^n}+\cdots +u_{2^{n+1}-1}& \le & 2^n{u}_{2^n}.\end{array}$$

   Then, denoting by $\left(s_n\right)$ the sequence of partial sums of ${\sum }_{n=1}^{+\infty }{u}_n$, we can write the following:

   $$\begin{array}{lll}{s}_n& =& \sum _{k=1}^n{u}_k\\ & =& u_1+u_2+u_3+\cdots +u_n\\ & \le & u_1+\left(u_2+u_3\right)+\left(u_4+u_5+u_6+u_7+u_8\right)+\cdots \\ & & +\left(u_{2^n}+u_{2^n+1}+\cdots +u_{2^{n+1}-1}\right)\\ & \le & u_1+2u_2+4u_4+8u_8+\cdots +2^n{u}_{2^n}\\ & =& \sum _{n=0}^{+\infty }{2}^n{u}_{2^n}.\end{array}$$

   We conclude that $\left(s_n\right)$ is bounded from above and, since it is an increasing sequence, we can conclude that it converges. Hence, the series ${\sum }_{n=1}^{+\infty }{u}_n$ is also convergent.

2. Suppose that the series ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent.

   Let $\left(s_n\right)$ denote the sequence of partial sums of the series ${\sum }_{n=0}^{+\infty }{2}^n{u}_{2^n}$. We have that

   $$\begin{array}{lll}\frac{1}{2}{s}_n& =& \frac{1}{2}\sum _{k=0}^n{2}^k{u}_{2^k}\\ & =& \frac{1}{2}\left(u_1+2u_2+4u_4+8u_8+\cdots +2^n{u}_{2^n}\right)\\ & =& \frac{1}{2}{u}_1+u_2+2u_4+4u_8+\cdots +2^{n-1}{u}_{2^n}\\ & \le & u_1+u_2+\left(u_3+u_4\right)+\left(u_5+u_6+u_7+u_8\right)+\cdots \\ & & +\left(u_{2^{n-1}+1}+u_{2^{n-1}+2}+\cdots +u_{2^n-1}+u_{2^n}\right)\\ & \le & \sum _{n=1}^{+\infty }{u}_n.\end{array}$$

   Therefore, we have that $s_n\le 2{\sum }_{n=1}^{+\infty }{u}_n$, which means that $\left(s_n\right)$ is bounded from above and, since it is an increasing sequence, it is convergent. Hence, ${\sum }_{n=0}^{+\infty }{2}^n{u}_{2^n}$ is also a convergent series.

□

We will now present a special kind of series.

Definition 3.4 (p-series). For any real number p, a p-series is a series defined as

$$\sum _{n=1}^{+\infty }\frac{1}{n^p}.$$

The convergence of a p-series is determined by the value of p as it is shown next.

Theorem 3.8. Let p be a real number. The series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$ is convergent if $p>1$ and divergent if $p\le 1$.

Proof. Let p be a real number and consider the p-series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}.$ The proof will be separated into different cases.

1. If $p=1$, then the p-series corresponds to the Harmonic series which we already know, from Example 3.9, that is divergent.

2. If $0<p<1$, then we consider the following trivial inequalities:

   $$n^p\le n\iff \frac{1}{n^p}\ge \frac{1}{n}.$$

   Therefore, since ${\sum }_{n=1}^{+\infty }\frac{1}{n}$ is divergent, see (1.), by the comparison test, we have that the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$ is also divergent.

3. If p is a natural number greater or equal to 2, consider the trivial inequalities

   $$n^p>n\left(n-1\right)\iff \frac{1}{n^p}<\frac{1}{n\left(n-1\right)}.$$

   Now, observe that

   $$\begin{array}{lll}\sum _{n=2}^{+\infty }\frac{1}{n\left(n-1\right)}& =& \sum _{n=2}^{+\infty }\left(\frac{-1}{n}+\frac{1}{n-1}\right)\\ & =& \sum _{n=2}^{+\infty }\left(\frac{1}{n-1}-\frac{1}{n}\right)\\ & =& \sum _{n=1}^{+\infty }\left(\frac{1}{n}-\frac{1}{n+1}\right),\end{array}$$

   which is a Mengoli’s series with sum 1. Then, since ${\sum }_{n=2}^{+\infty }\frac{1}{n\left(n-1\right)}$ is convergent, by the comparison test, we have that ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$ is a convergent series.

4. If p is a real number greater or equal than 2 and not natural, choose k to be the greatest natural number smaller or equal than p. We have that $n^k<n^p$ and, therefore,

   $$\frac{1}{n^k}>\frac{1}{n^p}.$$

   Since the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^k}$ is convergent, see 4., applying the comparison test, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$ is also convergent.

5. If p is a real number such that $p\in ]1,2[$, consider the sequence $u_n$ whose general term is $u_n=\frac{1}{n^p}$. Then, we have

   $$\begin{array}{lll}{2}^n{a}_{2^n}& =& 2^n\frac{1}{{\left(2^n\right)}^p}\\ & =& \frac{1}{2^{\mathit{np}-n}}\\ & =& \frac{1}{2^{n\left(p-1\right)}}\\ & =& {\left(\frac{1}{2^{p-1}}\right)}^n.\end{array}$$

   Since $p\in ]1,2[$, we have that

   $$0<\frac{1}{2^{p-1}}<1$$

   and, therefore, the series ${\left(\frac{1}{2^{p-1}}\right)}^n$ is convergent, since it is a geometric series with ratio $\frac{1}{2^{p-1}}<1$. Finally, by the Cauchy condensation test, we have that the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$ is convergent

6. If $p=0,$ then ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}={\sum }_{n=1}^{+\infty }1$ which is a divergent series because $lim1=1\ne 0$ (Theorem 3.4).

7. If $p<0,$ then ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}={\sum }_{n=1}^{+\infty }{n}^{\mid p\mid },$ which is a divergent series because $lim{n}^{\mid p\mid }=+\infty \ne 0$ (Theorem 3.4).

□

Example 3.12. In this example, we will show that the series

$$\sum _{n=1}^{+\infty }\frac{n}{7^n}$$

converges using the comparison test.

As a preliminary observation, since $7^n$ is much bigger than n, as n grows, it is natural to expect that the series will converge. Let us see that, indeed, this happens.

Observe that:

$$\begin{array}{lll}{7}^n& =& {\left(1+6\right)}^n\\ & =& \sum _{j=0}^n\left(\begin{array}{l}n\\ j\end{array}\right)1^{n-j}{6}^j\\ & =& \left(\begin{array}{c}n\\ 0\end{array}\right)1^n{6}^0+\left(\begin{array}{c}n\\ 1\end{array}\right)1^{n-1}{6}^1+\left(\begin{array}{l}n\\ 2\end{array}\right)1^{n-2}{6}^2+\cdots +\left(\begin{array}{l}n\\ n\end{array}\right)1^0{6}^n\\ & >& \left(\begin{array}{l}n\\ 3\end{array}\right)1^{n-3}{6}^3\\ & =& \frac{n!}{3!\left(n-3\right)!}216\\ & =& \frac{216n\left(n-1\right)\left(n-2\right)}{3!}\\ & =& 36n\left(n-1\right)\left(n-2\right)\\ & >& 36{\left(n-2\right)}^2.\end{array}$$

Then, we also have that

$$\frac{1}{7^n}<\frac{1}{36{\left(n-2\right)}^2},$$

and

$$\frac{n}{7^n}<\frac{n}{36n{\left(n-2\right)}^2},$$

for all $n\ge 3$.

Now, since the p-series ${\sum }_{n=1}^{+\infty }\frac{1}{n^2}$ converges, by Theorem 3.8, then the series ${\sum }_{n=3}^{+\infty }\frac{1}{{\left(n-2\right)}^2}$ and $\frac{1}{36}{\sum }_{n=3}^{+\infty }\frac{1}{{\left(n-2\right)}^2}$ also converge.

Finally, by the comparison test, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{n}{7^n}$ is convergent.

In view of the fact that we have a good understanding of their natures, the geometric series and the p-series are of great importance in the context of the usage of the comparison test to study the convergence of many series.

We will now present another test that is called the limit comparison test.

Theorem 3.9 (Limit comparison test). Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of non-negative terms and ${\sum }_{n=1}^{+\infty }{v}_n$ be a series of positive terms. If $lim\frac{u_n}{v_n}$ exists, then the following statements hold.

1. If $lim\frac{u_n}{v_n}=r,$ with $r\in {\mathbb{R}}^{+},$ then ${\sum }_{n=1}^{+\infty }{u}_n$ converges (diverges) if and only if ${\sum }_{n=1}^{+\infty }{v}_n$ converges (diverges).

2. If $lim\frac{u_n}{v_n}=0,$ then:

   1. if ${\sum }_{n=1}^{+\infty }{v}_n$ converges, then ${\sum }_{n=1}^{+\infty }{u}_n$ converges.

   2. if ${\sum }_{n=1}^{+\infty }{u}_n$ diverges, then ${\sum }_{n=1}^{+\infty }{v}_n$ diverges.

3. If $lim\frac{u_n}{v_n}=+\infty ,$ then:

   1. if ${\sum }_{n=1}^{+\infty }{v}_n$ diverges, then ${\sum }_{n=1}^{+\infty }{u}_n$ diverges.

   2. if ${\sum }_{n=1}^{+\infty }{u}_n$ converges, then ${\sum }_{n=1}^{+\infty }{v}_n$ converges.

Proof. Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of non-negative terms and ${\sum }_{n=1}^{+\infty }{v}_n$ be a series of positive terms.

1. Suppose that $lim\frac{u_n}{v_n}=r$, with $r\in {\mathbb{R}}^{+}$. Then, there is a $m\in \mathbb{N}$ such that:

   $$n>m\Rightarrow r-\frac{r}{2}<\frac{u_n}{v_n}<r+\frac{r}{2}.$$

   Therefore, for $n>m$ we have the following inequalities

   $$\frac{u_n}{v_n}>\frac{r}{2}\iff u_n>\frac{r}{2}{v}_n$$E21

   and

   $$\frac{u_n}{v_n}<\frac{3r}{2}\iff u_n<\frac{3r}{2}{v}_n.$$E22

   If ${\sum }_{n=1}^{+\infty }{u}_n$ converges, so does the series ${\sum }_{n=m+1}^{+\infty }{u}_n$. Then, considering inequality (21) and applying the comparison test, we conclude that the series ${\sum }_{n=m+1}^{+\infty }\frac{r}{2}{v}_n$ is convergent and, therefore, ${\sum }_{n=1}^{+\infty }{v}_n$ is also convergent.

   If ${\sum }_{n=1}^{+\infty }{v}_n$ is convergent, ${\sum }_{n=m+1}^{+\infty }\frac{3r}{2}{v}_n$ is also convergent. Then, taking into consideration the inequality (22) and applying the comparison test, we conclude that ${\sum }_{n=m+1}^{+\infty }{u}_n$ converges and so does the series ${\sum }_{n=1}^{+\infty }{u}_n$.

2. Suppose that $lim\frac{u_n}{v_n}=0$. Then, there is a $m\in \mathbb{N}$ such that:

   $$n>m\Rightarrow -1<\frac{u_n}{v_n}<1.$$

   This means that, for $n>m$ we have that $u_n<v_n$.

   If ${\sum }_{n=1}^{+\infty }{v}_n$ converges, so does the series ${\sum }_{n=m+1}^{+\infty }{v}_n$. Then, applying the comparison test, we conclude that ${\sum }_{n=m+1}^{+\infty }{u}_n$ is convergent and, therefore, ${\sum }_{n=1}^{+\infty }{u}_n$ is also convergent. This proves assertion (i.). Assertion (ii.) is just the contrapositive implication of (i.), which is logically equivalent.

3. Suppose that $lim\frac{u_n}{v_n}=+\infty$. Then, there is a $m\in \mathbb{N}$ such that:

$$n>m\Rightarrow u_n>v_n.$$

If ${\sum }_{n=1}^{+\infty }{v}_n$ is divergent, ${\sum }_{n=m+1}^{+\infty }{v}_n$ is also divergent. Then, by the comparison test, we conclude that ${\sum }_{n=m+1}^{+\infty }{u}_n$ is divergent. Therefore, the series ${\sum }_{n=1}^{+\infty }{u}_n$ is also divergent. This proves assertion (i.). Assertion (ii.) is just the contrapositive implication of (i.), which is logically equivalent.

□

Much of the analysis of a series convergence is done through the medium of the comparison test or the limit comparison test together with the support of series whose nature we are familiar with, such as the geometric series or the p-series. Next, we present some examples.

Example 3.13. In this example, we will show that the series

$$\sum _{n=1}^{+\infty }\frac{n+1}{n^3+n^2+n+1}$$

is convergent.

As a preliminary observation, note that if n is sufficiently big, we have that

$$\frac{n+1}{n^3+n^2+n+1}\approx \frac{n}{n^3}=\frac{1}{n^2}.$$

It is, therefore, natural to compare the series ${\sum }_{n=1}^{+\infty }\frac{n+1}{n^3+n^2+n+1}$ to the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^2}$ which is a convergent p-series, since $p=2>1$.

Considering $u_n=\frac{n+1}{n^3+n^2+n+1}$ and $v_n=\frac{1}{n^2}$, we have:

$$\begin{array}{lll}lim\frac{u_n}{v_n}& =& \underset{n\to +\infty }{lim}\frac{\frac{n+1}{n^3+n^2+n+1}}{\frac{1}{n^2}}\\ & =& lim\frac{n^3+n^2}{n^3+n^2+n+1}\\ & =& lim\frac{n^3\left(1+\frac{1}{n}\right)}{n^3\left(1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}\right)}\\ & =& lim\frac{1+\frac{1}{n}}{1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}}\\ & =& \frac{1+0}{1+0+0+0}\\ & =& 1>0.\end{array}$$

Then, applying the limit comparison test, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^2}$ and ${\sum }_{n=1}^{+\infty }\frac{n+1}{n^3+n^2+n+1}$ have the same nature of convergence and, therefore, the series ${\sum }_{n=1}^{+\infty }\frac{n+1}{n^3+n^2+n+1}$ is convergent.

Example 3.14. Let’s show that the series

$$\sum _{n=1}^{+\infty }\frac{n}{n^2+n+1}$$

is divergent.

Consider the sequence $\left(u_n\right)$ whose general term is

$$u_n=\frac{n}{n^2+n+1}$$

and the sequence $\left(v_n\right)$ defined by $v_n=\frac{1}{n}$. Then, we have:

$$\begin{array}{lll}lim\frac{u_n}{v_n}& =& lim\frac{\frac{n}{n^2+n+1}}{\frac{1}{n}}\\ & =& \underset{n\to +\infty }{lim}\frac{n^2}{n^2+n+1}\\ & =& \underset{n\to +\infty }{lim}\frac{n^2}{n^2\left(1+\frac{1}{n}+\frac{1}{n^2}\right)}\\ & =& \underset{n\to +\infty }{lim}\frac{1}{1+\frac{1}{n}+\frac{1}{n^2}}\\ & =& \frac{1}{1+0+0}\\ & =& 1>0.\end{array}$$

Now, since the series ${\sum }_{n=1}^{+\infty }\frac{1}{n}$ diverges, it’s the harmonic series, then, by the limit comparison test, we conclude that ${\sum }_{n=1}^{+\infty }\frac{n}{n^2+n+1}$ is a divergent series.

Example 3.15. In this example, we will again apply the limit comparison test to show that the series

$$\sum _{n=1}^{+\infty }\frac{n}{2^n\left(n^5+1\right)}$$

is convergent.

In this case, we consider the sequences $\left(u_n\right)$ defined by

$$u_n=\frac{n}{2^n\left(n^5+1\right)}$$

and $\left(v_n\right)$ defined by $v_n=\frac{1}{n^4}$. Then, we have:

$$\begin{array}{lll}lim\frac{u_n}{v_n}& =& lim\frac{\frac{n}{2^n\left(n^5+1\right)}}{\frac{1}{n^4}}\\ & =& lim\frac{n^5}{2^n\left(n^5+1\right)}\\ & =& lim\frac{n^5}{n^5{2}^n\left(1+\frac{1}{n^5}\right)}\\ & =& lim\frac{1}{2^n\left(1+\frac{1}{n^5}\right)}\\ & =& \frac{1}{+\infty }\\ & =& 0.\end{array}$$

Since ${\sum }_{n=1}^{+\infty }\frac{1}{n^4}$ is a convergent series, because it is a p-series with $p=4>1$, then, by the limit comparison test, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{n}{2^n\left(n^5+1\right)}$ is also convergent.

Note that we could have also used the comparison test to prove the convergence of ${\sum }_{n=1}^{+\infty }\frac{n}{2^n\left(n^5+1\right)}$. Indeed, since we have, for all $n\in \mathbb{N}$, that

$$\begin{array}{ll}& 2^n\left(n^5+1\right)>n^5\\ \iff & \frac{1}{2^n\left(n^5+1\right)}<\frac{1}{n^5}\\ \iff & \frac{n}{2^n\left(n^5+1\right)}<\frac{n}{n^5}\\ \iff & \frac{n}{2^n\left(n^5+1\right)}<\frac{1}{n^4}\end{array}$$

and since ${\sum }_{n=1}^{+\infty }\frac{1}{n^4}$ is convergent, as we already observed, then, by the comparison test, we have that ${\sum }_{n=1}^{+\infty }\frac{n}{2^n\left(n^5+1\right)}$ is convergent.

We will now present another test for verifying the convergence of series of positive numbers.

Theorem 3.10 (Ratio test). Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of positive terms. If $lim\frac{u_{n+1}}{u_n}$ exists, then the following statements hold.

1. If $lim\frac{u_{n+1}}{u_n}$ is a real number smaller than 1, then ${\sum }_{n=1}^{+\infty }{u}_n$ is a convergent series.

2. If $lim\frac{u_{n+1}}{u_n}$ is a real number greater than 1, then ${\sum }_{n=1}^{+\infty }{u}_n$ is a divergent series.

3. If $lim\frac{u_{n+1}}{u_n}=1^{+}$, then ${\sum }_{n=1}^{+\infty }{u}_n$ is a divergent series.

4. If $lim\frac{u_{n+1}}{u_n}=1^{-}$, then no conclusion can be drawn about the convergence of ${\sum }_{n=1}^{+\infty }{u}_n$.

5. If $lim\frac{u_{n+1}}{u_n}=+\infty$, then ${\sum }_{n=1}^{+\infty }{u}_n$ is a divergent series.

Proof. Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of positive terms.

1. Let $lim\frac{u_{n+1}}{u_n}=r$, with r being a real number smaller than 1. Then, there is a $m\in \mathbb{N}$ such that

   $$n>m\Rightarrow \frac{u_{n+1}}{u_n}<r-\frac{r-1}{2}=\frac{r+1}{2}.$$E23

   Since $0<r<1$, we have that $\frac{r+1}{2}<1$. Let $s=\frac{r+1}{2}$. We can, then, rewrite (23) as

   $$n>m\Rightarrow \frac{u_{n+1}}{u_n}<s,$$

   which means that, for $n>m$, we have:

   $$\frac{u_{m+2}}{u_{m+1}}<s\iff u_{m+2}<u_{m+1}s.$$

   We can, therefore, deduce the following inequalities:

   $$\begin{array}{lll}{u}_{m+2}& <& u_{m+1}s\\ u_{m+3}& <& u_{m+2}s<u_{m+1}{s}^2\\ u_{m+4}& <& u_{m+3}s<u_{m+1}{s}^3\\ & ⋮& \\ u_{m+k+1}& <& u_{m+k}s<u_{m+1}{s}^k\end{array}$$

   Hence, we can conclude that:

   $$u_n<u_{m+1}{s}^{n-m-1}=\frac{u_{m+1}}{s^{m+1}}{s}^n,$$

   for all $n\ge m+2$.

   The series ${\sum }_{n=m+2}^{+\infty }\frac{u_{m+1}}{s^{m+1}}{s}^n$ is a convergent series since it is a geometric series with ratio s and such that $\mid s\mid <1$. Then, by the comparison test, we conclude that the series ${\sum }_{n=m+1}^{+\infty }{u}_n$ is convergent and, therefore, the series ${\sum }_{n=1}^{+\infty }{u}_n$ is also convergent.

2. Let $lim\frac{u_{n+1}}{u_n}=r$, with r being a real number bigger than 1. Then, there is a $m\in \mathbb{N}$ such that

   $$n>m\Rightarrow \frac{u_{n+1}}{u_n}>r-\frac{r-1}{2}=\frac{r+1}{2}.$$E24

   Since $r>1$, we have that $\frac{r+1}{2}>1$. Let $s=\frac{r+1}{2}>1$. Then, we can rewrite (24) as

   $$n>m\Rightarrow \frac{u_{n+1}}{u_n}>s,$$

   which means that, for $n>m$, we have

   $$\frac{u_{n+1}}{u_n}>s\iff u_{n+1}>{\mathit{su}}_n.$$

   We can, therefore, conclude that, for $n>m$, we have $u_{n+1}>u_n$. But then, the following statements are true:

   $$\begin{array}{lll}{u}_{m+2}& >& u_{m+1}\\ u_{m+3}& >& u_{m+2}>u_{m+1}\\ & ⋮& \\ u_{m+k+1}& >& u_{m+k}>\cdots >u_{m+1}.\end{array}$$

   Hence, $u_{m+k+1}>u_{m+1},\forall k\in \mathbb{N}.$ Then $u_n>u_{m+1}>0$ for all $n\ge m+2.$ But then $lim{u}_n>0$ and, therefore, the series ${\sum }_{n=1}^{+\infty }{u}_n$ diverges.

3. Let $lim\frac{u_{n+1}}{u_n}=1^{+}$. Then, there exists a $m\in \mathbb{N}$ such that for $n>m$ we have

   $$\frac{u_{n+1}}{u_n}>1.$$

   But that means that $u_n>u_m,\forall n>m.$ Then, $lim{u}_n\ne 0$ and, therefore, the series ${\sum }_{n=1}^{+\infty }{u}_n$ is divergent.

4. In the case that $lim\frac{u_{n+1}}{u_n}=1^{-}$ nothing can be concluded about the convergence of the series ${\sum }_{n=1}^{+\infty }{u}_n$. To show this, we will present two examples.

   On the one hand, if we consider the divergent harmonic series ${\sum }_{n=1}^{+\infty }\frac{1}{n}$, we have that $u_n=\frac{1}{n}$ and

   $$\begin{array}{lll}lim\frac{u_{n+1}}{u_n}& =& =lim\frac{\frac{1}{n+1}}{\frac{1}{n}}\\ & =& lim\frac{n}{n+1}\\ & =& lim\frac{n}{n\left(1+\frac{1}{n}\right)}\\ & =& lim\frac{1}{1+\frac{1}{n}}\\ & =& 1^{-}.\end{array}$$

   On another hand, if we consider the convergent series ${\sum }_{n=1}^{+\infty }\frac{1}{n^2}$, we have $u_n=\frac{1}{n^2}$ and

   $$\begin{array}{lll}lim\frac{u_{n+1}}{u_n}& =& lim\frac{\frac{1}{{\left(n+1\right)}^2}}{\frac{1}{n^2}}\\ & =& lim\frac{n^2}{{\left(n+1\right)}^2}\\ & =& lim\frac{n^2}{n^2\left(1+\frac{2}{n}+\frac{1}{n^2}\right)}\\ & =& lim\frac{1}{1+\frac{2}{n}+\frac{1}{n^2}}\\ & =& 1^{-}.\end{array}$$

   Thus, nothing can be concluded if $lim\frac{u_{n+1}}{u_n}=1^{-}$.

5. Let $lim\frac{u_{n+1}}{u_n}=+\infty$. Then, there is a natural number m such that

   $$n>p\Rightarrow \frac{u_{n+1}}{u_n}>1.$$

   Therefore, we have that $u_{n+1}>u_n$ for $n>m$. But this means that $u_{m+2}>u_{m+1},u_{m+3}>$ $u_{m+2},\dots ,u_{m+k+1}>u_{m+k},\forall k\in \mathbb{N},$ and, therefore, we conclude that $u_n>u_{m+1}$ for $n>m+1.$ Thus,

   $$lim{u}_n\ge u_{m+1}>0.$$

   Hence, $lim{u}_n\ne 0$ and ${\sum }_{n=1}^{+\infty }{u}_n$ is divergent.

□

The ratio test is often referred to as the d’Alembert criterion in honor of the French mathematician Jean-Baptiste le Rond d’Alembert, who is credited for publishing it first. Some applications are presented next.

Example 3.16. In this example, we will study the convergence of the series

$$\sum _{n=1}^{+\infty }\frac{n^3}{5^n}.$$

We start by considering $\left(u_n\right)$ to be the sequence whose general term is

$$u_n=\frac{n^3}{5^n}.$$

Now, applying the ratio test, we have

$$\begin{array}{lll}lim\frac{u_{n+1}}{u_n}& =& =lim\frac{\frac{{\left(n+1\right)}^3}{5^{n+1}}}{\frac{n^3}{5^n}}\\ & =& lim\frac{{\left(n+1\right)}^3}{5n^3}\\ & =& lim\frac{n^3{\left(1+\frac{1}{n}\right)}^3}{5n^3}\\ & =& lim\frac{{\left(1+\frac{1}{n}\right)}^3}{5}\\ & =& \frac{1}{5}<1.\end{array}$$

Therefore, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{n^3}{5^n}$ converges.

Example 3.17. In this example, we will again apply the ratio test in order to prove that the series

$$\sum _{n=1}^{+\infty }\frac{n!}{n^2}$$

is divergent.

Let $\left(u_n\right)$ be the sequence defined by

$$u_n=\frac{n!}{n^2}.$$

Then, we have

$$\begin{array}{lll}lim\frac{u_{n+1}}{u_n}& =& =lim\frac{\frac{\left(n+1\right)!}{{\left(n+1\right)}^2}}{\frac{n!}{n^2}}\\ & =& lim\left[\frac{\left(n+1\right)!}{n!}\frac{n^2}{{\left(n+1\right)}^2}\right]\\ & =& lim\left[\left(n+1\right)\frac{n^2}{n^2{\left(1+\frac{1}{n}\right)}^2}\right]\\ & =& lim\left[\left(n+1\right)\frac{1}{{\left(1+\frac{1}{n}\right)}^2}\right]\\ & =& \left(+\infty \right)\times 1\\ & =& +\infty .\end{array}$$

Then, by the ratio test, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{n!}{n^2}$ diverges.

Example 3.18. In this example, we will use the ratio test to show that

$$lim\frac{n}{2^n}=0.$$

Let $\left(u_n\right)$ be the sequence defined by

$$u_n=\frac{n}{2^n}.$$

Then, we have

$$\begin{array}{lll}lim\frac{u_{n+1}}{u_n}& =& lim\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}\\ & =& lim\frac{n+1}{n}\frac{2^n}{2^{n+1}}\\ & =& lim\frac{n+1}{n}\frac{1}{2}\\ & =& \frac{1}{2}lim\frac{n\left(1+\frac{1}{n}\right)}{n}\\ & =& \frac{1}{2}lim\frac{1+\frac{1}{n}}{1}\\ & =& \frac{1}{2}lim\left(1+\frac{1}{n}\right)\\ & =& \frac{1}{2}<1.\end{array}$$

Therefore, by the ratio test, we conclude that the series ${\sum }_{n=1}^n\frac{n}{2^n}$ converges and, therefore, we have that $lim\frac{n}{2^n}=0$.

We will now present another test for the convergence of sequences that is known as Cauchy’s root test or simply the root test.

Theorem 3.11 (Root test). Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of non-negative terms such that there is an order from which all terms of the series are positive. If $lim\sqrt[n]{u_n}$ exists, then the following statements are true.

1. If $lim\sqrt[n]{u_n}$ is a real number smaller than 1, then ${\sum }_{n=1}^{+\infty }{u}_n$ is a convergent series.

2. If $lim\sqrt[n]{u_n}$ is a real number greater than 1, then ${\sum }_{n=1}^{+\infty }{u}_n$ is a divergent series.

3. If $lim\sqrt[n]{u_n}=1^{+},$ then ${\sum }_{n=1}^{+\infty }{u}_n$ is a divergent series.

4. If $lim\sqrt[n]{u_n}=1^{-}$, then nothing can be concluded about the convergence of ${\sum }_{n=1}^{+\infty }{u}_n$.

5. If $lim\sqrt[n]{u_n}=+\infty$, then ${\sum }_{n=1}^{+\infty }{u}_n$ is a divergent series.

Proof. Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of non-negative terms such that there is an order from which all terms of the series are positive.

1. Let $lim\sqrt[n]{u_n}=r,$ where r is a real number smaller than 1. Then, there is a natural number m such that

   $$n>m\Rightarrow \sqrt[n]{u_n}<r-\frac{r-1}{2}=\frac{r+1}{2}.$$

   Let $s=\frac{r+1}{2}$. note that s is such that $0<s<1$. Then, for $n>m$, we have

   $$\sqrt[n]{u_n}<s\iff u_n<s^n.$$

   Hence, we have the following inequalities:

   $$\begin{array}{lll}{u}_{m+1}& <& s^{m+1}\\ u_{m+2}& <& s^{m+2}\\ & ⋮& \\ u_{m+k}& <& s^{m+k},\forall k\in \mathbb{N}\\ u_n& <& s^n,\forall n\ge m+1.\end{array}$$

   Now, since the series ${\sum }_{n=1}^{+\infty }{s}^n$ converges, as it is a geometric series with a ratio smaller than 1 in absolute value, then the series ${\sum }_{n=m+1}^{+\infty }{s}^n$ also converges. Then, by the comparison test, we conclude that the series ${\sum }_{n=m+1}^{+\infty }{u}_n$ is convergent and, therefore, the series ${\sum }_{n=1}^{+\infty }{u}_n$ is also convergent.

2. Let $lim\sqrt[n]{u_n}=r,$ where r is a real number bigger than 1. Then, there is a natural number m such that

   $$n>m\Rightarrow \sqrt[n]{u_n}>r-\frac{r-1}{2}=\frac{r+1}{2}.$$

   Let $s=\frac{r+1}{2}$. Note that s is such that $1<s<\sqrt[n]{u_n}$. Therefore, we have, for $n>m$, that $u_n>s^n.$ But then we have $u_n>1$ for $n>m$ and, therefore, $lim{u}_n\ge 1$, that is, $lim{u}_n\ne 0,$ which, in turn, implies that the series ${\sum }_{n=1}^{+\infty }{u}_n$ is divergent.

3. Let $lim\sqrt[n]{u_n}=1^{+}$. Then, there is a natural number m such that

   $$n>m\Rightarrow \sqrt[n]{u_n}\ge 1.$$

   Hence, for $n>m$, we have that $u_n>1$. Then, we can conclude that $lim{u}_n\ne 0$ and, therefore, the series ${\sum }_{n=1}^{+\infty }{u}_n$ is divergent.

4. In the case that $lim\sqrt[n]{u_n}=1^{-}$ nothing can be concluded about the convergence of the series ${\sum }_{n=1}^{+\infty }{u}_n$. To show this, we will present two examples.

   We first consider the series

   $$\sum _{n=1}^{+\infty }{\left(\frac{n}{n+1}\right)}^n.$$

   In this case, $\left(u_n\right)$ is the sequence defined by

   $$u_n={\left(\frac{n}{n+1}\right)}^n.$$

   The series ${\sum }_{n=1}^{+\infty }{\left(\frac{n}{n+1}\right)}^n$ is divergent since we have

   $$\begin{array}{lll}lim{u}_n& =& lim{\left(\frac{n}{n+1}\right)}^n\\ & =& lim\frac{n^n}{{\left(n+1\right)}^n}\\ & =& lim\frac{n^n}{n^n{\left(1+\frac{1}{n}\right)}^n}\\ & =& lim\frac{1}{{\left(1+\frac{1}{n}\right)}^n}\\ & =& \frac{1}{lim{\left(1+\frac{1}{n}\right)}^n}\\ & =& \frac{1}{e}\ne 0.\end{array}$$

   Yet, the root test yields:

   $$\begin{array}{lll}lim\sqrt[n]{u_n}& =& lim\sqrt[n]{{\left(\frac{n}{n+1}\right)}^n}\\ & =& lim\frac{n}{n+1}\\ & =& lim\frac{n}{n\left(1+\frac{1}{n}\right)}\\ & =& lim\frac{1}{\left(1+\frac{1}{n}\right)}\\ & =& 1^{-}.\end{array}$$

   Now, we consider the series

   $$\sum _{n=1}^{+\infty }\frac{n+1}{n^3}.$$

   In this case, $\left(u_n\right)$ is the sequence whose general term is

   $$u_n=\frac{n+1}{n^3}.$$

   In the case of this series, the root test also yields $1^{-}$:

   $$\begin{array}{lll}lim\sqrt[n]{u_n}& =& lim\sqrt[n]{\frac{n+1}{n^3}}\\ & =& lim\frac{\sqrt[n]{n+1}}{\sqrt[n]{n^3}}\\ & =& lim\frac{\sqrt[n]{n+1}}{{\left(\sqrt[n]{n}\right)}^3}\\ & =& lim\frac{\sqrt[n]{n}\sqrt[n]{1+\frac{1}{n}}}{\sqrt[n]{n}{\left(\sqrt[n]{n}\right)}^2}\\ & =& lim\frac{\sqrt[n]{1+\frac{1}{n}}}{{\left(\sqrt[n]{n}\right)}^2}\\ & =& 1^{-}.\end{array}$$

   But the series ${\sum }_{n=1}^{+\infty }\frac{n+1}{n^3}$ is a convergent series. Indeed, considering $\left(v_n\right)$ to be the sequence defined by $v_n=\frac{1}{n^2}$, we have

   $$\begin{array}{lll}lim\frac{u_n}{v_n}& =& lim\frac{\frac{n+1}{n^3}}{\frac{1}{n^2}}\\ & =& lim\frac{n^2\left(n+1\right)}{n^3}\\ & =& lim\frac{n^3\left(1+\frac{1}{n}\right)}{n^3}\\ & =& lim\left(1+\frac{1}{n}\right)\\ & =& 1\ne 0.\end{array}$$

   Since the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^2}$ is convergent, it is a p-series with $p=2>1$, then, by the limit comparison test, we conclude that ${\sum }_{n=1}^{+\infty }\frac{n+1}{n^3}$ is also a convergent series.

   Thus, nothing can be concluded if $lim\sqrt[n]{u_n}=1^{-}$.

5. The proof of this assertion is analogous to the one produced for (b).

□

Next, we present a couple of examples where we apply the root test.

Example 3.19. Consider the series

$$\sum _{n=1}^{+\infty }{\left(\frac{1+2n}{n}\right)}^n.$$

Let $\left(u_n\right)$ be the sequence defined by

$$u_n={\left(\frac{1+2n}{n}\right)}^n.$$

Then, we have

$$\begin{array}{lll}lim\sqrt[n]{u_n}& =& lim\sqrt[n]{{\left(\frac{1+2n}{n}\right)}^n}\\ & =& lim\frac{1+2n}{n}\\ & =& lim\frac{n\left(\frac{1}{n}+2\right)}{n}\\ & =& lim\frac{\left(\frac{1}{n}+2\right)}{1}\\ & =& lim\left(\frac{1}{n}+2\right)\\ & =& 2>1.\end{array}$$

Then, by the root test, we conclude that ${\sum }_{n=1}^{+\infty }{\left(\frac{1+2n}{n}\right)}^n$ is a divergent series.

Example 3.20. Consider the series

$$\sum _{n=3}^{+\infty }\frac{1}{{\left(lnn\right)}^n}.$$

Let $\left(u_n\right)$ be the sequence whose general term is

$$u_n=\frac{1}{{\left(lnn\right)}^n}.$$

Then, we have

$$\begin{array}{lll}lim\sqrt[n]{u_n}& =& lim\sqrt[n]{\frac{1}{{\left(lnn\right)}^n}}\\ & =& lim\frac{1}{lnn}\\ & =& 0<1.\end{array}$$

Then, by the root test, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{1}{{\left(lnn\right)}^n}$ is convergent.

It can be proved that the root test is stronger than the previously presented ratio test in the sense that whenever the ratio test is conclusive, so is the root test, but the converse is not true.

Proposition 3.3. Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of positive terms. If the ratio test proves that ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent, then the root test also proves that ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent.

Proof. Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of positive terms.

If the ratio test proves that ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent, then we have that $lim\frac{u_{n+1}}{u_n}=r$, where r is a real number smaller than 1. Then, there is a natural m such that:

$$n>m\Rightarrow \frac{u_{n+1}}{u_n}<r-\frac{r-1}{2}=\frac{r+1}{2}.$$

Let $s=\frac{r+1}{2}$. Note that $0<s<1$. Then, for $n>m$, we have:

$$\frac{u_{n+1}}{u_n}<s\iff u_{n+1}<u_{n}s.$$

We can, therefore, write the following inequalities

$$\begin{array}{lll}{u}_{m+2}& <& u_{m+1}s\\ u_{m+3}& <& u_{m+2}s<u_{m+1}{s}^2\\ & ⋮& \\ u_{m+k+1}& <& u_{m+1}{s}^k,\forall k\in \mathbb{N}.\end{array}$$

This means that, for $n>m$, we have

$$u_n<u_{m+1}{s}^{n-m-1}.$$E25

From (25), we can deduce the following:

$$\begin{array}{lll}\sqrt[n]{u_n}& <& \sqrt[n]{u_{m+1}{s}^{n-m-1}}\\ & =& \sqrt[n]{u_{m+1}}\times s^{\frac{n-m-1}{n}}\\ & =& \sqrt[n]{u_{m+1}}\times s^{1-\frac{m+1}{n}}.\end{array}$$

Applying limits to both sides of the inequality $\sqrt[n]{u_n}<\sqrt[n]{u_{m+1}}\times s^{1-\frac{m+1}{n}}$, we obtain

$$\begin{array}{lll}lim\sqrt[n]{u_n}& <& lim\left(\sqrt[n]{u_{m+1}}\times s^{1-\frac{m+1}{n}}\right)\\ & =& 1\times s^{1-0}\\ & =& s.\end{array}$$

We can, therefore, conclude that $lim\sqrt[n]{u_n}<1$ which means that the root test also proves that ${\sum }_{n=1}^{+\infty }{u}_n$ is a convergent series. □

Proposition 3.3 shows that whenever the ratio test provides information on a series convergence, the root test will also provide that information. Additionally, there are some examples of series for which we can determine its convergence with the root test but not with the ratio test. We present one of those next.

Example 3.21. Consider the series ${\sum }_{n=1}^{+\infty }{u}_n$, where $\left(u_n\right)$ is the sequence defined by

$$u_n=\left\{\begin{array}{cc}\frac{1}{3^{n+1}}& ,\mathit{\text{if}}n\mathit{\text{is}}\mathit{odd}\\ \frac{1}{3^n}& ,\mathit{\text{if}}n\mathit{\text{is\hspace{0.17em}\hspace{0.17em}even}}.\end{array}\right.$$

If we apply the ratio test, we obtain

$$\begin{array}{lll}lim\frac{u_{n+1}}{u_n}& =& lim\frac{\frac{1}{3^{n+1}}}{\frac{1}{3^n}}\\ & =& lim\frac{3^n}{3^{n+1}}\\ & =& \frac{1}{3}\end{array}$$

if n is even and

$$\begin{array}{lll}lim\frac{u_{n+1}}{u_n}& =& lim\frac{\frac{1}{3^n}}{\frac{1}{3^{n+1}}}\\ & =& lim\frac{3^{n+1}}{3^n}\\ & =& 3\end{array}$$

if n is odd, which means that the limit does not exist rendering the test inconclusive.

However, if we apply the root test, we obtain

$$\begin{array}{lll}lim\sqrt[n]{u_n}& =& lim\sqrt[n]{\frac{1}{3^n}}\\ & =& \frac{1}{3}\end{array}$$

if n is even and

$$\begin{array}{lll}lim\sqrt[n]{u_n}& =& lim\sqrt[n]{\frac{1}{3^{n+1}}}\\ & =& lim\frac{1}{\sqrt[n]{3^{n+1}}}\\ & =& lim\frac{1}{3^{\frac{n+1}{n}}}\\ & =& lim\frac{1}{3^{1+\frac{1}{n}}}\\ & =& lim\frac{1}{3\times 3^{\frac{1}{n}}}\\ & =& lim\frac{1}{3\times 1}\\ & =& \frac{1}{3}\end{array}$$

if n is odd. Thus, $lim\sqrt[n]{u_n}=\frac{1}{3}<1$ and we conclude that the series ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent.

Note that it is not our intention, with Proposition 3.3 and Example 3.21, to suggest that the ratio test is useless. Despite the root test being a stronger test, the ratio test, as some of our previous examples show, proves to be, in many situations, easier to apply than the root test. Thus, its pertinence is justified.

We will now present the last test in this section that can be applied to series of positive terms.

Theorem 3.12 (Integral test). Let $f:[1,+\infty [\to {\mathbb{R}}^{+}$ be a monotonically decreasing and integrable function and $\left(u_n\right)$ be a sequence such that $u_n=f\left(n\right),\forall n\in \mathbb{N}$. Then, the series ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent if and only if the improper integral ${\int }_1^{+\infty }f\left(x\right)\mathit{dx}$ converges.

Proof. Let $f:[1,+\infty [\to {\mathbb{R}}^{+}$ be a monotonically decreasing function and $\left(u_n\right)$ be a sequence such that $u_n=f\left(n\right),\forall n\in \mathbb{N}$. Also, let $\left(s_n\right)$ denote the sequence of partial sums of ${\sum }_{n=1}^{+\infty }{u}_n$ and $\left(v_n\right)$ denote the sequence whose general term is

$$v_n={\int }_1^{n}f\left(x\right)\mathit{dx}.$$

Before we start the proof itself, we need to deduce some preliminary relations between the terms of the sequences $\left(s_n\right)$ and $\left(v_n\right)$. Since f is a decreasing function, we have, for any natural m, that

$$f\left(m+1\right)\le {\int }_m^{m+1}f\left(x\right)\mathit{dx}\le f\left(m\right)$$

which can be rewritten as:

$$u_{m+1}\le {\int }_m^{m+1}f\left(x\right)\mathit{dx}\le u_m.$$E26

Applying the principle of (26) recursively, we obtain:

$$u_2+u_3+\cdots +u_n\le v_n\le u_1+u_2+\cdots +u_{n-1}.$$E27

From (27), we conclude two things: that

$$v_n\le s_{n-1}$$E28

and that

$$\begin{array}{ll}& u_2+u_3+\cdots +u_n\le v_n\\ \iff & u_1+u_2+u_3+\cdots +u_n\le u_1+v_n\\ \iff & s_n\le u_1+v_n.\end{array}$$E29

We are now in conditions to prove Theorem 3.12. Suppose that ${\sum }_{n=1}^{+\infty }{u}_n$ is a convergent series. Then, $\left(s_n\right)$ is a convergent sequence. Since $\left(s_n\right)$ converges, then, by (28), the sequence $\left(v_n\right)$ is bounded from above. Note that $\left(v_n\right)$ is an increasing sequence because f is a positive function. Since $\left(v_n\right)$ is increasing and is bounded from above we can conclude that $\left(v_n\right)$ is also a convergent sequence. Thus, the improper integral ${\int }_1^{+\infty }f\left(x\right)\mathit{dx}$ converges.

Conversely, suppose that ${\int }_1^{+\infty }f\left(x\right)\mathit{dx}$ converges. Then, $\left(v_n\right)$ is a convergent sequence. Then, by (29), we conclude that $\left(s_n\right)$ is bounded from above and, therefore, since it is an increasing sequence, it is also convergent. □

Note that the integral test can be straightforwardly generalized for functions defined in the interval $[a,+\infty [,\forall a\in \mathbb{N}$. The integral test proves to be very useful for series whose general term converts immediately into a decreasing and positive function defined in the interval $[1,+\infty [$ whose primitive is known. For instance, the integral test can be used to easily determine the nature of convergence of the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$, with $p\in \mathbb{R},\frac{1}{nln\left(n\right)}$ or ${\sum }_{n=1}^{+\infty }\frac{1}{n^2+1}$, as we will see in the following examples, with which we finish this section.

Example 3.22. Let us show, using the integral test, that the p-series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$, with $p\in \mathbb{R}$, is convergent if $p>1$ and divergent if $0<p\le 1$.

Let p be a real number greater than 0 and f the function defined by

$$f\left(x\right)=\frac{1}{x^p}.$$

We start by observing that f is a continuous, positive, and decreasing function in the interval $[1,+\infty [$. Indeed, since

$$\begin{array}{lll}{f}^{\prime }\left(x\right)& =& {\left(x^{-p}\right)}^{\prime }\\ & =& -{\mathit{px}}^{-p-1}\\ & =& -\frac{p}{x^{p+1}}\end{array}$$

then we conclude that $f^{\prime }\left(x\right)<0,\forall x\in [1,+\infty [$, and, therefore, f is decreasing on the interval $[1,+\infty [$.

Suppose that $p>1$. Then, we have

$$\begin{array}{lll}{\int }_1^{+\infty }f\left(x\right)\mathit{dx}& =& \underset{b\to +\infty }{lim}{\int }_1^b\frac{1}{x^p}\mathit{dx}\\ & =& \underset{b\to +\infty }{lim}{\int }_1^b{x}^{-p}\mathit{dx}\\ & =& \underset{b\to +\infty }{lim}{\left[\frac{x^{-p+1}}{-p+1}\right]}_1^b\\ & =& \underset{b\to +\infty }{lim}\left[\frac{1}{\left(-p+1\right)b^{p-1}}-\frac{1}{-p+1}\right]\\ & =& 0+\frac{1}{p-1}\\ & =& \frac{1}{p-1}.\end{array}$$

Therefore, we conclude that the improper integral ${\int }_1^{+\infty }f\left(x\right)\mathit{dx}$ is convergent and, by the integral test, the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$ is convergent when $p>1$.

Now, suppose that p is such that $0<p<1$. In this case, we have

$$\begin{array}{lll}{\int }_1^{+\infty }f\left(x\right)\mathit{dx}& =& \underset{b\to +\infty }{lim}{\int }_1^b\frac{1}{x^p}\mathit{dx}\\ & =& \underset{b\to +\infty }{lim}{\int }_1^b{x}^{-p}\mathit{dx}\\ & =& \underset{b\to +\infty }{lim}{\left[\frac{x^{-p+1}}{-p+1}\right]}_1^b\\ & =& \underset{b\to +\infty }{lim}\left[\frac{1}{\left(-p+1\right)b^{p-1}}-\frac{1}{-p+1}\right]\\ & =& +\infty +\frac{1}{p-1}\\ & =& +\infty .\end{array}$$

So, in this case, the improper integral ${\int }_1^{+\infty }f\left(x\right)\mathit{dx}$ diverges and, therefore, the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$, with $p\in ]0,1[$, is divergent by the integral test.

Finally, let’s suppose that $p=1$. In this case we have

$$\begin{array}{lll}{\int }_1^{+\infty }f\left(x\right)\mathit{dx}& =& \underset{b\to +\infty }{lim}{\int }_1^{b}f\left(x\right)\mathit{dx}\\ & =& \underset{b\to +\infty }{lim}{\int }_1^b\frac{1}{x}\mathit{dx}\\ & =& \underset{b\to +\infty }{lim}{\left[ln\left(x\right)\right]}_1^b\\ & =& \underset{b\to +\infty }{lim}\left(ln\left(b\right)-ln\left(1\right)\right)\\ & =& \underset{b\to +\infty }{lim}ln\left(b\right)\\ & =& +\infty .\end{array}$$

Similarly to the last case studied, the improper integral ${\int }_1^{+\infty }f\left(x\right)\mathit{dx}$ diverges and, therefore, the integral test allows us to conclude that the series ${\sum }_{n=1}^{+\infty }\frac{1}{n^p}$, with $p=1$, is divergent.

Example 3.23. In this last example, we will analyze the convergence of the series

$$\sum _{n=1}^{+\infty }\frac{1}{\left(n+2\right)ln\left(n+2\right)}$$

using the integral test.

We start by considering the function f defined by

$$f\left(x\right)=\frac{1}{\left(x+2\right)ln\left(x+2\right)},$$

for $x\in [1,+\infty [$.

Note that $f\left(x\right)>0,\forall x\in [1,+\infty [$. Additionally, we have

$$\begin{array}{lll}{f}^{\prime }\left(x\right)& =& {\left({\left(\left(x+2\right)ln\left(x+2\right)\right)}^{-1}\right)}^{\prime }\\ & =& \left(-1\right)({\left(\left(x+2\right)ln\left(x+2\right)\right)}^{-2}\left(ln\left(x+2\right)+\frac{x+2}{x+2}\right)\\ & \text{=}& \left(-1\right)({\left(\left(x+2\right)ln\left(x+2\right)\right)}^{-2}\left(ln\left(x+2\right)+1\right)\end{array}$$

and, therefore, $f^{\prime }\left(x\right)<0,\forall x\in [1,+\infty [$ and f is a decreasing function in the interval $[1,+\infty [.$

Now, since

$$\begin{array}{lll}{\int }_1^{+\infty }\frac{1}{\left(x+2\right)ln\left(x+2\right)}\mathit{dx}& =& \underset{b\to +\infty }{lim}{\int }_1^b\frac{\frac{1}{x+2}}{ln\left(x+2\right)}\mathit{dx}\\ & =& {\left[ln\left(ln\left(x+2\right)\right)\right]}_1^b\\ & =& \underset{b\to +\infty }{lim}\left[ln\left(ln\left(b+2\right)\right)-ln\left(ln\left(3\right)\right)\right]\\ & =& +\infty -ln\left(\mathit{ln}\left(3\right)\right)\\ & =& +\infty ,\end{array}$$

we have that the improper integral ${\int }_1^{+\infty }\frac{1}{\left(x+2\right)ln\left(x+2\right)}\mathit{dx}$ diverges and, therefore, by the integral test, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{1}{\left(n+2\right)ln\left(n+2\right)}$ is divergent.

3.3 Alternating series

In this section, we will focus on the study of the convergence of alternating series. Informally, an alternating series is just a series whose consecutive terms have opposite signs. The formal definition is as follows.

Definition 3.5 (Alternating series). An alternating series is a series of the form

$$\sum _{n=1}^{+\infty }{\left(-1\right)}^{n-1}{u}_n,$$

where $\left(u_n\right)$ is a sequence of positive terms.

The study of the convergence of alternating series often starts with a test most commonly known as the alternating series test.

Theorem 3.13 (Alternating series test). Let ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^{n-1}{u}_n$, with $\left(u_n\right)$ being a sequence of positive terms, be an alternating series. If $\left(u_n\right)$ is a decreasing sequence and $lim{u}_n=0$, then the alternating series ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^{n-1}{u}_n$ is convergent.

Proof. Let $\left(u_n\right)$ be a decreasing sequence of positive terms such that:

$$lim{u}_n=0.$$

Since $\left(u_n\right)$ is decreasing, we have that $u_n>u_{n+1},\forall n\in \mathbb{N}$. Let $\left(s_n\right)$ denote the sequence of partial sums of the series ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^{n-1}{u}_n$. Now, consider the sequence $\left(s_{2n}\right)$. We have:

$$\begin{array}{lll}{s}_2& =& u_1-u_2;\\ s_4& =& \left(u_1-u_2\right)+\left(u_3-u_4\right)\\ s_6& =& \left(u_1-u_2\right)+\left(u_3-u_4\right)+\left(u_5-u_6\right)\\ & ⋮& \\ s_{2n}& =& \left(u_1-u_2\right)+\left(u_3-u_4\right)+\left(u_5-u_6\right)+\cdots +\left(u_{2n-1}-u_{2n}\right)\end{array}$$

and, therefore, we conclude that

$$s_2<s_4<s_6<\cdots <s_{2n-2}<s_{2n},$$

that is, $\left(s_{2n}\right)$ is an increasing sequence.

Additionally, we have

$$\begin{array}{lll}{s}_2& =& u_1-u_2<u_1\\ s_4& =& u_1-\left(u_2-u_3\right)-u_4<u_1\\ s_6& =& u_1-\left(u_2-u_3\right)-\left(u_4-u_5\right)-u_6<u_1\\ & ⋮& \\ s_{2n}& =& u_1-\left(u_2-u_3\right)-\left(u_4-u_5\right)-\cdots -\left(u_{2n-2}-u_{2n-1}\right)-u_{2n})<u_1.\end{array}$$

This means that $s_{2n}<u_1$, for all $n\in \mathbb{N}$, and, therefore, we conclude that $s_{2n}$ is a convergent sequence since it is an increasing sequence bounded from above by $u_1$.

Now observe that, since

$$s_{2n}=\left(u_1-u_2\right)+\left(u_3-u_4\right)+\left(u_5-u_6\right)+\cdots +\left(u_{2n-1}-u_{2n}\right)$$

and

$$s_{2n-1}=\left(u_1-u_2\right)+\left(u_3-u_4\right)+\left(u_5-u_6\right)+\cdots +\left(u_{2n-3}-u_{2n-2}\right)+u_{2n-1},$$

we have that $s_{2n}=s_{2n-1}-u_{2n}$. Then, we have

$$lim{s}_{2n}=lim{s}_{2n-1}-lim{u}_{2n}$$

and since $lim{u}_n=0$ we conclude that

$$lim{s}_{2n}=lim{s}_{2n-1}.$$

We can, therefore, conclude that the sequence of partial sums $\left(s_n\right)$ converges and, thus, the series ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^{n-1}{u}_n$ is convergent. □

The alternating series test is normally credited to the philosopher and mathematician Gottfried Leibniz and, therefore, it is frequently referred to as Leibniz’s test, Leibniz’s rule, or even the Leibniz criterion.

When dealing with an alternating series, it is a good strategy to study the series of the absolute values of the terms as there is a relationship between the convergence of these two series. That relationship is observed in the next result.

Theorem 3.14. If the series ${\sum }_{n=1}^{+\infty }\mid u_n\mid$ is convergent, then the series ${\sum }_{n=1}^{+\infty }{u}_n$ is convergent.

Proof. Let ${\sum }_{n=1}^{+\infty }\mid u_n\mid$ be a convergent series.

Firstly, observe that

$$\begin{array}{ll}& -\mid u_n\mid \le u_n\le \mid u_n\mid \\ \iff & 0\le u_n+\mid u_n\mid \le 2\mid u_n\mid ,\end{array}$$

for all $n\in \mathbb{N}$.

Now, note that ${\sum }_{n=1}^{+\infty }\left(2|u_n|\right)$ is a convergent series due to our hypothesis that the series ${\sum }_{n=1}^{+\infty }\mid u_n\mid$ converges.

Then, by the comparison test, we have that the series ${\sum }_{n=1}^{+\infty }\left(u_n+|u_n|\right)$ is convergent.

Finally, observe that the series ${\sum }_{n=1}^{+\infty }{u}_n$ can be written as

$$\begin{array}{lll}\sum _{n=1}^{+\infty }{u}_n& =& \sum _{n=1}^{+\infty }\left(u_n+|u_n|-|u_n|\right)\\ & =& \sum _{n=1}^{+\infty }\left(u_n+|u_n|\right)-\sum _{n=1}^{+\infty }\mid u_n\mid .\end{array}$$

Then, since ${\sum }_{n=1}^{+\infty }{u}_n$ is equal to the difference between two convergent series, we conclude that it is also convergent. □

Given a certain series, Theorem 3.14 exhibits the importance of the convergence of the series of the absolute values of the terms to conclude on the convergence of the original series. Also, since the result contains only one implication and not an equivalence, it points out to a certain division in the series’ universe between the convergent series whose series of absolute values also converge and those convergent series whose series of absolute values do not converge. Such a division is clarified with the next definition.

Definition 3.6 (Absolute and conditional convergence). A series ${\sum }_{n=1}^{+\infty }{u}_n$ is said to be:

1. absolutely convergent if ${\sum }_{n=1}^{+\infty }\mid u_n\mid$ converges.

2. conditionally convergent if ${\sum }_{n=1}^{+\infty }{u}_n$ converges but ${\sum }_{n=1}^{+\infty }\mid u_n\mid$ diverges.

Next, we present a couple of examples.

Example 3.24. Consider the series

$$\sum _{n=1}^{+\infty }{\left(-1\right)}^n\frac{1}{n}.$$

Let $\left(u_n\right)$ be the sequence whose general term is $u_n=\frac{1}{n}$. We have that $\left(u_n\right)$ is a decreasing sequence of positive terms and that $lim{u}_n=0$. Then, by the alternating series test, the series ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^n{u}_n$ converges.

The corresponding series of absolute values of ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^n\frac{1}{n}$ is the series ${\sum }_{n=1}^{+\infty }\frac{1}{n}$ which is the already studied divergent harmonic series.

Therefore, the series ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^n\frac{1}{n}$ is conditionally convergent.

Example 3.25. Consider the series

$$\sum _{n=1}^{+\infty }\frac{sin\left(\mathit{n\pi }\right)}{n^4}.$$

Note that

$$\left|\frac{sin\left(\mathit{n\pi }\right)}{n^4}\right|\le \frac{1}{n^4},$$

for all $n\in \mathbb{N}$, and that the series

$$\sum _{n=1}^{+\infty }\frac{1}{n^4}$$

is convergent because it is a p-series with $p=4>1$. Then, by the comparison test, we conclude that the series

$$\sum _{n=1}^{+\infty }\left|\frac{sin\left(\mathit{n\pi }\right)}{n^4}\right|$$

is convergent and, therefore, we can conclude that the series

$$\sum _{n=1}^{+\infty }\frac{sin\left(\mathit{n\pi }\right)}{n^4}$$

is absolutely convergent.

The ratio and root tests presented in Section 3.2 for series of non-negative terms can be straightforwardly adapted for series with terms with variable sign, including series of alternating terms, as we present next.

Theorem 3.15 (Generalized ratio test). Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series such that there is a $m\in \mathbb{N}$ for which we have $u_n\ne 0$ for all $n>m$. If $lim\frac{\mid u_{n+1}\mid }{\mid u_n\mid }$ exists, then the following statements are true.

1. If $lim\frac{\mid u_{n+1}\mid }{\mid u_n\mid }$ is a real number smaller than 1, then the series ${\sum }_{n=1}^{+\infty }{u}_n$ is absolutely convergent.

2. If $lim\frac{\mid u_{n+1}\mid }{\mid u_n\mid }$ is a real number greater than 1, then the series ${\sum }_{n=1}^{+\infty }\mid u_n\mid$ and ${\sum }_{n=1}^{+\infty }{u}_n$ are divergent.

3. If $lim\frac{\mid u_{n+1}\mid }{\mid u_n\mid }=1^{+}$, then the series ${\sum }_{n=1}^{+\infty }{u}_n$ is divergent.

4. If $lim\frac{\mid u_{n+1}\mid }{\mid u_n\mid }=1^{-},$ then nothing can be concluded about the convergence of the series ${\sum }_{n=1}^{+\infty }\mid u_n\mid$ and ${\sum }_{n=1}^{+\infty }{u}_n$.

5. If $lim\frac{\mid u_{n+1}\mid }{\mid u_n\mid }=+\infty$, then the series ${\sum }_{n=1}^{+\infty }\mid u_n\mid$ and ${\sum }_{n=1}^{+\infty }{u}_n$ are divergent.

Theorem 3.16 (Generalized root test). Let ${\sum }_{n=1}^{+\infty }{u}_n$ be a series of real numbers. If $lim\sqrt[n]{\mid u_n\mid }$ exists, then the following statements are true.

1. If $lim\sqrt[n]{\mid u_n\mid }$ is a real number smaller than 1, then ${\sum }_{n=1}^{+\infty }{u}_n$ is absolutely convergent.

2. If $lim\sqrt[n]{\mid u_n\mid }$ is a real number greater than 1, then ${\sum }_{n=1}^{+\infty }{u}_n$ is divergent.

3. If $lim\sqrt[n]{\mid u_n\mid }=1^{+},$ then ${\sum }_{n=1}^{+\infty }{u}_n$ is divergent.

4. If $lim\sqrt[n]{\mid u_n\mid }=1^{-},$ nothing can be concluded about the convergence of the series ${\sum }_{n=1}^{+\infty }{u}_n$.

5. If $lim\sqrt[n]{\mid u_n\mid }=+\infty ,$ then ${\sum }_{n=1}^{+\infty }{u}_n$ is divergent.

In the following paragraphs we present two examples with which we finish this section.

Example 3.26. In this example we will study the convergence of the series

$$\sum _{n=1}^{+\infty }\frac{n!}{2^n}{x}^n,$$

where x is a real number.

Let $\left(u_n\left(x\right)\right)$ be the sequence whose general term is given by

$$u_n\left(x\right)=\left|\frac{n!}{2^n}{x}^n\right|.$$

Applying the ratio test, we have, for $x\ne 0$,

$$\begin{array}{lll}lim\frac{u_{n+1}\left(x\right)}{u_n\left(x\right)}& =& lim\frac{\left|\frac{\left(n+1\right)!}{2^{n+1}}{x}^{n+1}\right|}{\left|\frac{n!}{2^n}{x}^n\right|}\\ & =& lim\left|\frac{\left(n+1\right)!}{2^{n+1}}{x}^{n+1}\frac{2^n}{n!x^n}\right|\\ & =& lim\left|\frac{x\left(n+1\right)}{2}\right|\\ & =& \left|\frac{x}{2}\right|lim\left(n+1\right)\\ & =& \left|\frac{x}{2}\right|\times \left(+\infty \right)\\ & =& +\infty \end{array}$$

and, therefore, we conclude that the series ${\sum }_{n=1}^{+\infty }\frac{n!}{2^n}{x}^n$ is divergent for $x\ne 0$.

If $x=0$, then clearly we have that the series ${\sum }_{n=1}^{+\infty }\frac{n!}{2^n}{x}^n$ is convergent with sum equal to 0, since it is an infinite sum of null terms.

Example 3.27. In this example we will study the convergence of the series

$$\sum _{n=1}^{+\infty }n{\left(x-1\right)}^n,$$

where x is a real number.

Let $\left(u_n\left(x\right)\right)$ be the sequence defined by $u_n\left(x\right)=\mid n{\left(x-1\right)}^n\mid$. Then, applying the root test, we have

$$\begin{array}{lll}lim\sqrt[n]{u_n\left(x\right)}& =& lim\sqrt[n]{n\mid {\left(x-1\right)}^n\mid }\\ & =& lim\sqrt[n]{n}\sqrt[n]{{\left|x-1\right|}^n}\\ & =& lim\sqrt[n]{n}\mid x-1\mid \\ & =& 1\times \mid x-1\mid \\ & =& \mid x-1\mid .\end{array}$$

Hence, we have:

• if $\mid x-1\mid <1$, that is, if $x\in ]0,2[,$ then ${\sum }_{n=1}^{+\infty }n{\left(x-1\right)}^n$ is absolutely convergent;

• if $\mid x-1\mid >1,$ that is, if $x\in ]-\infty ,0\left[\cup \right]2,+\infty [,$ then ${\sum }_{n=1}^{+\infty }n{\left(x-1\right)}^n$ diverges.

For $x=0$ then, we have

$$\sum _{n=1}^{+\infty }n{\left(x-1\right)}^n=\sum _{n=1}^{+\infty }n{\left(-1\right)}^n.$$

In this case, let $\left(u_n\right)$ be the sequence defined by $u_n=n{\left(-1\right)}^n$. Now, if we consider the sequences of even and odd terms of $\left(u_n\right),\left(u_{2n}\right)$ and $\left(u_{2n-1}\right)$, respectively, then their general terms are $u_{2n}=2n$ and $u_{2n-1}=1-2n$ and we have that $lim{u}_{2n}=+\infty$ and $lim{u}_{2n-1}=-\infty$. Therefore, ${lim}_{n\to +\infty }{u}_n$ does not exist and, consequently, the series ${\sum }_{n=1}^{+\infty }n{\left(-1\right)}^n$ is divergent.

For $x=2$ we have that

$$\sum _{n=1}^{+\infty }n{\left(x-1\right)}^n=\sum _{n=1}^{+\infty }n$$

which is divergent since ${lim}_{n\to +\infty }n=+\infty$.

Therefore, the series ${\sum }_{n=1}^{+\infty }n{\left(x-1\right)}^n$ is convergent if and only if $x\in ]0,2[$.

3.4 Function sequences and series

In this section, we aim to study the convergence of a special type of series whose terms are functions themselves. These series have a great importance in several branches of Mathematics such as differential equations or approximation theory. Before, however, we must also study briefly the corresponding sequences of functions which are the generating sequences of those series.

3.4.1 Sequences of functions

A sequence of functions is just a sequence whose terms are functions instead of numbers.

Definition 3.7 (Sequence of functions). A set ${\left(f_n\right)}_{n\in \mathbb{N}}$, or simply $\left(f_n\right)$, is called a sequence of functions if, for each natural n, $f_n$ is a function $f_n:A\to \mathbb{R}$, with A, the common domain of all terms of $\left(f_n\right)$, being a subset of $\mathbb{R}$.

Unlike sequences of real numbers, a sequence of functions can converge in different manners. The first type of convergence, somewhat the most natural one, is called pointwise convergence, and it is presented next.

Definition 3.8 (Pointwise convergence). Let $\left(f_n\right)$ be a sequence of functions with common domain $A\subset \mathbb{R}$. Then, $\left(f_n\right)$ converges pointwise to the function $f:A\to \mathbb{R}$ if, for all $x\in A$, we have:

$$\underset{n\to +\infty }{lim}{f}_n\left(x\right)=f\left(x\right).$$

In the study of sequences of functions, we deal with two variables: n, the natural variable of the sequence, and x, the real variable of the functions, which are the terms of the sequences. In view of this and to avoid confusion, along this section, we will write ${lim}_{n\to +\infty }{f}_n\left(x\right)$ and not $lim{f}_n\left(x\right),$ although all limits will be evaluated when n tends to $+\infty$.

An example of pointwise convergence is presented next.

Example 3.28. Let $\left(f_n\right)$ be the sequence of functions defined as

$$\begin{array}{ccc}{f}_n:\left[0,1\right]& \to & \mathbb{R}\\ x& \mapsto & x^n\end{array}.$$

The sequence of functions $\left(f_n\right)$ converges pointwise to the function

$$f\left(x\right)=\left\{\begin{array}{lll}0& \mathit{\text{if}}& x\ne 1\\ 1& \mathit{\text{if}}& x=1\end{array}\right..$$

Indeed, we have

$$\underset{n\to +\infty }{lim}{f}_n\left(x\right)=\underset{n\to +\infty }{lim}{x}^n=0=f\left(x\right)$$

if $x\in [0,1[$ and

$$\underset{n\to +\infty }{lim}{f}_n\left(1\right)=1=f\left(1\right).$$

Saying that a sequence of functions $\left(f_n\right)$ converges pointwise to a function f, that is, that ${lim}_{n\to +\infty }{f}_n\left(x\right)=f\left(x\right),\forall x\in A$, is equivalent to say that

$$\forall x\in A,\forall \epsilon >0,\exists m\in \mathbb{N}:\forall n\in \mathbb{N}:\left(n\ge m\right)\Rightarrow \mid f_n\left(x\right)-f\left(x\right)\mid <\epsilon .$$

This means that, for each x and $\epsilon$, we can find an order m from which $f_n\left(x\right)$ differs from f(x) no more than $\epsilon$.

A stronger form of convergence is presented next.

Definition 3.9 (Uniform convergence). Let $\left(f_n\right)$ be a sequence of functions with common domain $A\subset \mathbb{R}$. Then, $\left(f_n\right)$ converges uniformly to the function $f:A\to \mathbb{R}$ if:

$$\forall \epsilon >0,\exists m\in \mathbb{N}:\forall n\in \mathbb{N},\forall x\in A,\left(n>m\right)\Rightarrow \mid f_n\left(x\right)-f\left(x\right)\mid <\epsilon .$$

Example 3.29. Let $\left(f_n\right)$ be the sequence of functions whose general term is

$$\begin{array}{ccc}{f}_n:A& \to & \mathbb{R}\\ x& \mapsto & \frac{x}{n}\end{array},$$

where $A\subset \mathbb{R}$.

The sequence $\left(f_n\right)$ converges pointwise to the function $f\left(x\right)=0$, since, for each $x\in \mathbb{R}$, we have:

$$\begin{array}{lll}\underset{n\to +\infty }{lim}{f}_n\left(x\right)& =& \underset{n\to +\infty }{lim}\frac{x}{n}\\ & =& \frac{x}{+\infty }\\ & =& 0.\end{array}$$

However, uniform convergence of $\left(f_n\right)$ is not guaranteed as it depends on the domain A.

If $A=\left[0,1\right]$, for instance, then the series of functions $\left(f_n\right)$ converges uniformly to the function

$$\begin{array}{ccc}f:\left[0,1\right]& \to & \mathbb{R}\\ x& \mapsto & 0\end{array}.$$

Indeed, let $\epsilon >0$. Then

$$\begin{array}{lll}\mid f_n\left(x\right)-f\left(x\right)\mid <\epsilon & \iff & \left|\frac{x}{n}-0\right|<\epsilon \\ & \iff & \frac{\mid x\mid }{n}<\epsilon \\ & \iff & n>\frac{\mid x\mid }{\epsilon }.\end{array}$$

Since $x\in \left[0,1\right]$, then

$$\frac{\mid x\mid }{\epsilon }\le \frac{1}{\epsilon }.$$

Therefore, choosing $m=\frac{1}{\epsilon }+1$, for example, we have that

$$\forall \epsilon >0,\exists m\in \mathbb{N}:\forall n\in \mathbb{N},\left(n>m\right)\Rightarrow \mid f_n\left(x\right)-f\left(x\right)\mid <\epsilon ,\forall x\in \left[0,1\right],$$

which means that $\left(f_n\right)$ is uniformly convergent.

If $A=\mathbb{R}$, nonetheless, it is easy to see that, given an $\epsilon >0$, we cannot find an m such that $\mid f_n\left(x\right)-f\left(x\right)\mid <\epsilon ,$ for each $x\in \mathbb{R}$. In fact, if $\epsilon =1$, we would have

$$\begin{array}{lll}\mid f_n\left(x\right)-f\left(x\right)\mid <1& \iff & \left|\frac{x}{n}-0\right|<1\\ & \iff & \frac{\mid x\mid }{n}<1\\ & \iff & n>\mid x\mid ,\end{array}$$

which means that there is not any order m such that $\mid f_n\left(x\right)-f\left(x\right)\mid <1$ for each $x\in \mathbb{R}$. Hence, $\left(f_n\right)$ does not converge uniformly, although it is pointwise convergent.

As it follows from the previous definitions, we have the following relationship between these two types of convergence of sequences of functions.

Proposition 3.4. Let $\left(f_n\right)$ be a sequence of functions. If $\left(f_n\right)$ is uniformly convergent, then $\left(f_n\right)$ is pointwise convergent.

Given a sequence of functions $\left(f_n\right)$ and a function f, all of them bounded functions on a domain A, consider the sequence of real numbers $\left(M_n\right)$ defined by:

$$M_n=\underset{x\in A}{sup}\mid f_n\left(x\right)-f\left(x\right)\mid .$$

Then, the sequence of functions $\left(f_n\right)$ converges uniformly to f if and only if $lim{M}_n=0$.

In the cases that, for all $n\in \mathbb{N},f_n$ and f are continuous functions and $A=\left[a,b\right],\left(M_n\right)$ can be defined by

$$M_n=\underset{x\in A}{max}\mid f_n\left(x\right)-f\left(x\right)\mid$$

and the sequence $\left(f_n\right)$ converges uniformly to the function f if and only if $lim{M}_n=0$.

Note that in Example 3.29, when $A=\left[0,1\right]$, we have

$$\begin{array}{lll}{M}_n& =& \underset{x\in \left[0,1\right]}{sup}\left|f_n\left(x\right)-f\left(x\right)\right|\\ & =& \underset{x\in \left[0,1\right]}{max}\left|f_n\left(x\right)-f\left(x\right)\right|\\ & =& \underset{x\in \left[0,1\right]}{max}\left|\frac{x}{n}\right|\\ & =& \frac{1}{n}\end{array}$$

and, therefore, since $lim{M}_n=0,$ the sequence $\left(f_n\right)$ converges uniformly to the function $f\left(x\right)=0.$

Example 3.30. Let $\left(f_n\right)$ be the sequence of functions defined by:

$$\begin{array}{ccc}{f}_n:[1,+\infty [& \to & \mathbb{R}\\ x& \mapsto & e^{-n^{2}x}\end{array}.$$

We will show that $\left(f_n\right)$ is uniformly convergent to the function f defined by

$$\begin{array}{lll}f:[1,+\infty [& \to & \mathbb{R}\\ x& \mapsto & 0\end{array}.$$

Indeed, we have that

$$\begin{array}{lll}{M}_n& =& \underset{x\in [1,+\infty [}{sup}\mid f_n\left(x\right)-f\left(x\right)\mid \\ & =& \underset{x\in [1,+\infty [}{sup}\mid e^{-n^{2}x}-0\mid \\ & =& \frac{1}{e^{n^2}}.\end{array}$$

Since

$$\begin{array}{lll}lim{M}_n& =& lim\frac{1}{e^{n^2}}\\ & =& \frac{1}{+\infty }\\ & =& 0,\end{array}$$

then the sequence of functions $\left(f_n\right)$ converges uniformly to the function f.

Note, however, that if the domain of the functions considered was the set $]0,+\infty [$, instead of $[1,+\infty [$, we would have

$$\begin{array}{lll}{M}_n& =& \underset{x\in ]0,+\infty [}{sup}\mid f_n\left(x\right)-f\left(x\right)\mid \\ & =& \underset{x\in ]0,+\infty [}{sup}\mid e^{-n^{2}x}-0\mid \\ & =& 1\end{array}$$

and, therefore, $lim{M}_n=1\ne 0$. Hence, $\left(f_n\right)$ would not be uniformly convergent.

The next result relates the convergence of a sequence of functions with the continuity of the functions involved.

Theorem 3.17. Let A be a subset of $\mathbb{R}$, a be an interior point of A and $\left(f_n\right)$ be a sequence of functions with domain A that is uniformly convergent to the function f also of domain A. Then, if for all $n\in \mathbb{N},f_n$ is a continuous function at a, then f is continuous at a.

Proof. Let A be a subset of $\mathbb{R}$, a be an interior point of A and $\left(f_n\right)$ be a sequence of functions with domain A that is uniformly convergent to the function f also of domain A.

Suppose that, for each natural n, $f_n$ is a continuous function at a. Since the sequence $\left(f_n\right)$ converges uniformly to f let m be the natural number such that for $n>m$ we have

$$\mid f_n\left(x\right)-f\left(x\right)\mid <\frac{\epsilon }{3},$$

$\forall x\in A$. Let $n=m+1$. The function $f_{m+1}$ is a continuous function at a by hypothesis. Then, let $\delta$ be a real positive number such that $]a-\delta ,a+\delta [\subset A$ and

$$\mid f_{m+1}\left(x\right)-f_{m+1}\left(a\right)\mid <\frac{\epsilon }{3},$$

$\forall x\in ]a-\delta ,a+\delta [.$ Then, for $x\in ]a-\delta ,a+\delta [,$ we have the following:

$$\begin{array}{lll}\mid f\left(x\right)-f\left(a\right)\mid & <& \mid f\left(x\right)-f_{m+1}\left(x\right)+f_{m+1}\left(x\right)-f_{m+1}\left(a\right)+f_{m+1}\left(a\right)-\\ & & -f\left(a\right)\mid \\ & \le & \mid f\left(x\right)-f_{m+1}\left(x\right)\mid +\mid f_{m+1}\left(x\right)-f_{m+1}\left(a\right)\mid +\\ & & +\mid f_{m+1}\left(a\right)-f\left(a\right)\mid \\ & \le & \frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}\\ & \le & \epsilon .\end{array}$$

Hence, we conclude that f is continuous at a. □

Example 3.31. Consider the sequence of functions $\left(f_n\right)$ such that

$$\begin{array}{lll}{f}_n:\left[0,1\right]& \to & \mathbb{R}\\ x& \mapsto & x^n\end{array},$$

$\forall n\in \mathbb{N}$, and the real function f also of domain [0, 1] defined by

$$f\left(x\right)=\left\{\begin{array}{ll}0,& \mathit{\text{if}}x\in [0,1[\\ 1,& \mathit{\text{if}}x=1\end{array}\right..$$

The sequence $\left(f_n\right)$ converges pointwise to the function f. Indeed, for $x\in [0,1[$, we have that

$$\underset{n\to +\infty }{lim}{f}_n\left(x\right)=\underset{n\to +\infty }{lim}{x}^n=0$$

and, for $x=1$, we have

$$\underset{n\to +\infty }{lim}{f}_n\left(1\right)=\underset{n\to +\infty }{lim}{1}^n=1.$$

However, by Theorem 3.17, $\left(f_n\right)$ does not converge uniformly to the function f since all the functions $f_n$ are continuous and f is a discontinuous function.

Theorem 3.17 and Example 3.31 provide a powerful criterium for a sequence of continuous functions not being uniformly convergent to a function. If that function is not continuous, then uniform convergence does not occur.

Example 3.32. Consider the sequence of functions $\left(f_n\right)$ such that

$$\begin{array}{lll}{f}_n:[0,+\infty [& \to & \mathbb{R}\\ x& \mapsto & \frac{x}{1+\mathit{nx}}\end{array},$$

$\forall n\in \mathbb{N}$. Let us show that $\left(f_n\right)$ converges uniformly to the function f of domain $[0,+\infty [$, such that $f\left(x\right)=0$.

Firstly, we note that

$$\begin{array}{lll}\underset{n\to +\infty }{lim}{f}_n\left(x\right)& =& \underset{n\to +\infty }{lim}\frac{x}{1+\mathit{nx}}\\ & =& \frac{x}{+\infty }\\ & =& 0.\end{array}$$

Now, since, for $x\ne 0$, we have

$$\frac{x}{1+\mathit{nx}}\le \frac{x}{\mathit{nx}},$$

then, for each $n\in \mathbb{N}$, we have

$$0\le f_n\left(x\right)\le \frac{1}{n}.$$

We can, therefore, write that

$$0\le \left|f_n\left(x\right)-f\left(x\right)\right|\le \frac{1}{n},$$

$\forall x\in [0,+\infty [$, which implies that

$$\forall \epsilon >0,\exists m=⌊\frac{1}{\epsilon }⌋+1,n>m\to \mid f_n\left(x\right)-f\left(x\right)\mid <\epsilon ,\forall x\in [0,+\infty [,$$

that is, $\left(f_n\right)$ converges uniformly to f.

Next, we present a necessary and sufficient condition for the uniform convergence of a sequence of functions $\left(f_n\right)$ known as the Cauchy criterion.

Theorem 3.18 (Cauchy criterion – version 1). The sequence of functions $\left(f_n\right)$, all of domain A, converges uniformly if and only if:

$$\forall \epsilon >0,\exists p\in \mathbb{N}:\left(n\ge p\wedge m\ge p\right)\Rightarrow \mid f_n\left(x\right)-f_m\left(x\right)\mid <\epsilon ,\forall x\in A.$$

Sometimes Theorem 3.18 is presented in the following manner.

Theorem 3.19 (Cauchy criterion – version 2). The sequence of functions $\left(f_n\right)$, all of domain A, converges uniformly if and only if:

$$\forall \epsilon >0,\exists m\in \mathbb{N}:\forall k\in \mathbb{N},\left(n>m\right)\Rightarrow \mid f_{n+k}\left(x\right)-f_n\left(x\right)\mid <\epsilon ,\forall x\in A.$$

Now, we present a result that deals with sequences of Riemann integrable functions. Informally, a Riemann integrable function is just a function for which we can approximate the area under its curve using step functions, that is, rectangles with an equal given length, and that approximation can be as tight as we want. The Riemann integral, named after the German mathematician Bernhard Riemann, is credited to be the first rigorous definition of the integral of a function on a given interval.

Theorem 3.20. Let A be a subset of $\mathbb{R}$ and $\left(g_n\right)$ a sequence of Riemann integral functions in $\left[a,b\right]\subset A$ converging uniformly to a function g of domain [a, b]. Then, g is integrable on [a, b] and we have the following equality:

$${\int }_a^{b}g\left(x\right)\mathit{dx}=\underset{n\to +\infty }{lim}{\int }_a^b{g}_n\left(x\right)\mathit{dx}.$$

Proof. Since it is out of the scope of our book and it requires some more advanced calculus tools, we will skip the proof that g is integrable.

Let A be a subset of $\mathbb{R}$ and $\left(g_n\right)$ a sequence of Riemann integral functions in $\left[a,b\right]\subset A$ converging uniformly to a function g of domain [a, b].

Let $\epsilon$ be a positive real number. Since $\left(g_n\right)$ converges uniformly to g, then let m be a natural number such that, for $n>m$ we have

$$\mid g_n\left(x\right)-g\left(x\right)\mid <\frac{\epsilon }{b-a},$$

$\forall x\in \left[a,b\right]$. But then we have

$$\begin{array}{lll}\left|{\int }_a^b{g}_n\left(x\right)\mathit{dx}-{\int }_a^{b}g\left(x\right)\mathit{dx}\right|& =& \left|{\int }_a^b\left(g_n\left(x\right)-g\left(x\right)\right)\mathit{dx}\right|\\ & \le & {\int }_a^b\mid g_n\left(x\right)-g\left(x\right)\mid \mathit{dx}\\ & \le & {\int }_a^b\frac{\epsilon }{b-a}\mathit{dx}\\ & \le & \frac{\epsilon }{b-a}{\int }_a^b\mathit{dx}\\ & \le & \frac{\epsilon }{b-a}{\left[x\right]}_a^b\mathit{dx}\\ & \le & \frac{\epsilon }{b-a}\left(b-a\right)\\ & \le & \epsilon .\end{array}$$

Since we proved that

$$\exists m\in \mathbb{N}:\left(n>m\right)\Rightarrow \left|{\int }_a^b{g}_n\mathit{dx}-{\int }_a^{b}g\left(x\right)\mathit{dx}\right|<\epsilon ,$$

we conclude that

$$\underset{n\to +\infty }{lim}{\int }_a^b{g}_n\left(x\right)\mathit{dx}={\int }_a^{b}g\left(x\right)\mathit{dx}.$$

□

We now present an example of an application of Theorem 3.13.

Example 3.33. In this example, we will show that the sequence of functions $\left(f_n\right)$ of domain [0, 1] defined by

$$f_n\left(x\right)=\left\{\begin{array}{lll}4n^{2}x& ,& \mathit{\text{if}}x\in \left[0,\frac{1}{4n}\right]\\ -4n^{2}x+2n& ,& \mathit{\text{if}}x\in \left]\frac{1}{4n},\frac{1}{2n}\right]\\ 0& ,& \mathit{\text{if}}x\in \left]\frac{1}{2n},1\right]\end{array}\right.,$$

converges pointwise to the null function, f, defined on [0, 1] but does not converge uniformly to f.

Let $x\in ]0,1]$. Next, we choose $m\in \mathbb{N}$ such that $m=⌊\frac{1}{2x}⌋+1$. Then, for $n>m$ we have $n>\frac{1}{2x}$, which means that, for $n>m$, we have $\frac{1}{2n}<x$. Therefore, for $n>m$, we have that $f_n\left(x\right)=0$ and, consequently,

$$\underset{n\to +\infty }{lim}{f}_n\left(x\right)=0=f\left(x\right).$$

Now, let $x=0$. Then, since $f_n\left(0\right)=0$, we have

$$\underset{n\to +\infty }{lim}{f}_n\left(0\right)=0=f\left(x\right)$$

and, from all of the above, we conclude that $\left(f_n\right)$ converges pointwise to f.

As for uniform convergence, observe that

$${\int }_0^1{f}_n\left(x\right)\mathit{dx}=\frac{1}{4}$$

and

$${\int }_0^{1}f\left(x\right)\mathit{dx}={\int }_0^{1}0\mathit{dx}=0.$$

Then, by Theorem 3.13, since

$$\underset{n\to +\infty }{lim}{\int }_0^1{f}_n\left(x\right)=\frac{1}{4}\ne 0={\int }_0^{1}f\left(x\right)\mathit{dx},$$

we conclude that $\left(f_n\right)$ does not converge uniformly to the null function f.

We finish this section with a result on sequences of differentiable functions.

Theorem 3.21. Let $\left(f_n\right)$ be a sequence of differentiable functions with domain [a, b]. If the sequence $\left(f_n^{\prime }\right)$, of the derivatives of $f_n$, converges uniformly to the function g of domain [a, b] and if there is a $c\in \left[a,b\right]$ such that $\left(f_n\left(c\right)\right)$ converges, then the sequence $\left(f_n\right)$ converges uniformly to a function f of domain [a, b] such that f is differentiable and, for all $x\in \left[a,b\right]$, we have that

$$f^{\prime }\left(x\right)=\underset{n\to +\infty }{lim}{f}_n\left(x\right)=g\left(x\right).$$

Proof. Let $x\in \left[a,b\right]$ and $c\in \left[a,b\right]$ such that the sequence $\left(f_n\left(c\right)\right)$ converges. By Theorem 3.13 we can write that

$${\int }_c^{x}g\left(u\right)\mathit{du}=\underset{n\to +\infty }{lim}{\int }_c^x{f}_n^{\prime }\left(u\right)\mathit{du}.$$

Since g is a continuous function, then we have

$${\int }_c^{x}g\left(u\right)\mathit{du}=\underset{n\to +\infty }{lim}\left(f_n\left(x\right)-f\left(c\right)\right).$$

Therefore, we conclude that $\left(f_n\right)$ converges to the function f defined by

$$f\left(x\right)={\int }_c^{x}g\left(u\right)\mathit{du}+f\left(c\right),$$

$\forall x\in \left[a,b\right]$. Since g is a continuous function, we can, therefore, conclude that $f^{\prime }\left(x\right)=g\left(x\right)$.

Now, since $\left(f_n^{\prime }\right)$ converges uniformly to g, let $m_1$ be a natural number such that, for $n>m_1,$ we have

$$\left|g\left(u\right)-f_n^{\prime }\left(u\right)\right|<\frac{\epsilon }{b-a}.$$

Since $\left(f_n\left(c\right)\right)$ converges pointwise to f(c), then let $m_2$ be a natural number such that, for $n>m_2,$ we have

$$\mid f_n\left(c\right)-f\left(c\right)\mid <\frac{\epsilon }{2}.$$

Then, for $n>m=max\left\{m_1,m_2\right\},$ we have

$$\begin{array}{lll}\mid f\left(x\right)-f_n\left(x\right)\mid & =& \left|{\int }_c^{x}g\left(u\right)\mathit{du}+f\left(c\right)-\left({\int }_a^x{f}_n^{\prime }\left(u\right)\mathit{du}+f_n\left(c\right)\right)\right|\\ & <& \left|\left({\int }_c^{x}g\left(u\right)\mathit{du}-{\int }_c^x{f}_n^{\prime }\left(u\right)\mathit{du}\right)+\left(f\left(c\right)-f_n\left(c\right)\right)\right|\\ & \le & \left|{\int }_c^{x}g\left(u\right)\mathit{du}-{\int }_c^x{f}_n^{\prime }\left(u\right)\mathit{du}\right|+\mid f\left(c\right)-f_n\left(c\right)\mid \\ & \le & {\int }_c^x\mid g\left(u\right)-f_n^{\prime }\left(u\right)\mid \mathit{du}+\mid f\left(c\right)-f_n\left(c\right)\mid \\ & \le & {\int }_c^x\frac{\epsilon }{2\left(b-a\right)}\mathit{du}+\frac{\epsilon }{2}\\ & \le & \frac{\epsilon }{2\left(b-a\right)}\left(x-c\right)+\frac{\epsilon }{2}\\ & \le & \frac{\epsilon }{2\left(b-a\right)}\left(b-a\right)+\frac{\epsilon }{2}\\ & \le & \frac{\epsilon }{2}+\frac{\epsilon }{2}\\ & \le & \frac{\epsilon }{2}.\end{array}$$

Therefore, for $n>m,$ we have that $\mid f\left(x\right)-f_n\left(x\right)\mid <\epsilon ,$ for all $x\in \left[a,b\right].$ Hence, $\left(f_n\right)$ converges uniformly to f. □

3.4.2 Function series

In the following, we define a function series and present some important results.

Definition 3.10 (Function series). A function series is an infinite sum ${\sum }_{n=1}^{+\infty }{f}_n$, where $\left(f_n\right)$ is a sequence of functions defined in a set $A\subset \mathbb{R}$.

As we did with series of real numbers, for each function series, we consider the correspondent sequence of partial sums of order n, $\left(S_n\right)$, defined by: $S_n={\sum }_{k=1}^n{f}_k$.

Now, we extend the concepts of pointwise and uniform convergence from sequences of functions to function series.

Definition 3.11 (Pointwise convergence). Let $\left(f_n\right)$ be a sequence of functions defined in a set $A\subset \mathbb{R}$ and $f:A\to \mathbb{R}$. The function series ${\sum }_{n=1}^{+\infty }{f}_n$ is said to be pointwise convergent to the sum function f in a set A, and we write ${\sum }_{n=1}^{+\infty }{f}_n=f,$ if the corresponding sequence of partial sums $\left(S_n\right)$ is pointwise convergent to f in A, that is, if $lim{S}_n\left(x\right)=f\left(x\right),\forall x\in A$.

Definition 3.12 (Uniform convergence). Let $\left(f_n\right)$ be a sequence of functions defined in a set $A\subset \mathbb{R}$ and $f:A\to \mathbb{R}$. The function series ${\sum }_{n=1}^{+\infty }{f}_n$ is said to be uniformly convergent to the function f in A if the corresponding sequence of partial sums $\left(S_n\right)$ is uniformly convergent to f.

An example is presented next.

Example 3.34. Let $\left(f_n\right)$ be the sequence of real functions with common domain $]-1,1[$ such that, for each $n\in \mathbb{N},f_n\left(x\right)=x^n$.

The corresponding sequence of partial sums is:

$$\begin{array}{lll}{S}_n\left(x\right)& =& \sum _{k=1}^n{x}^{k-1}\\ & =& 1+x+x^2+\cdots +x^{n-1}\\ & =& \frac{1-x^n}{1-x}.\end{array}$$

Hence, for $x\in ]-1,1[$, we have

$$\begin{array}{lll}S\left(x\right)& =& lim{S}_n\left(x\right)\\ & =& lim\frac{1-x^n}{1-x}\\ & =& \frac{1-lim{x}^n}{1-x}\\ & =& \frac{1-0}{1-x}\\ & =& \frac{1}{1-x}.\end{array}$$

Therefore, we conclude that the series ${\sum }_{n=1}^{+\infty }{f}_n$ is pointwise convergent to the function:

$$\begin{array}{ccc}f:]-1,1[& \to & \mathbb{R}\\ x& \mapsto & \frac{1}{1-x}\end{array}.$$

Definition 3.13 (Domain of convergence). Let $\left(f_n\right)$ be a sequence of functions of domain $A\subset \mathbb{R}.$ The set of all elements $x\in A$ for which the series of real numbers ${\sum }_{n=1}^{+\infty }{f}_n\left(x\right)$ converges is called the domain of convergence.

Example 3.35. Consider the series of functions defined by

$$\sum _{n=1}^{+\infty }{3}^{n}sin\left(\frac{x}{5^n}\right).$$

We start by observing that:

$$\begin{array}{lll}\left|3^{n}sin\left(\frac{x}{5^n}\right)\right|& \le & \mid 3^n\mid \left|sin\left(\frac{x}{5^n}\right)\right|\\ & \le & 3^n\left|\frac{x}{5^n}\right|\\ & \le & \mid x\mid \left(\frac{3^n}{5^n}\right)\\ & \le & \mid x\mid {\left(\frac{3}{5}\right)}^n,\end{array}$$

for all $x\in \mathbb{R}$ and for all $n\in \mathbb{N}$.

Now, since the series ${\sum }_{n=1}^{+\infty }\mid x\mid {\left(\frac{3}{5}\right)}^n$ converges for all $x\in \mathbb{R},$ then the series ${\sum }_{n=1}^{+\infty }{3}^{n}sin\left(\frac{x}{5^n}\right)$ is absolutely convergent for any $x\in \mathbb{R}$. Therefore, the domain of convergence of the function series ${\sum }_{n=1}^{+\infty }{3}^{n}sin\left(\frac{x}{5^n}\right)$ is $\mathbb{R}$.

We continue this survey by presenting two relevant results for determining uniform convergence of function series.

Theorem 3.22 (Cauchy’s uniform convergence criterion for series of functions). Let $\left(f_n\right)$ be a sequence of functions with common domain $A\subset \mathbb{R}$. Then, the series of functions ${\sum }_{n=1}^{+\infty }{f}_n$ is uniformly convergent in A if

$$\forall \epsilon >0,\exists p\in \mathbb{N}:\forall m\in \mathbb{N},\forall x\in A,\left(k>p\right)\Rightarrow \left|\sum _{i=k+1}^{k+m}{f}_i\left(x\right)\right|<\epsilon .$$

Theorem 3.23 (Weierstrass M-Test). Let $\left(f_n\right)$ be a sequence of functions with common domain $A\subset \mathbb{R}$. Then, if there is a sequence $\left(M_n\right)$ of real non-negative numbers such that

1. ${\sum }_{n=1}^{+\infty }{M}_n$ is convergent and

2. $\forall n\in \mathbb{N},\forall x\in A,\mid f_n\left(x\right)\mid \le M_n,$

the series of functions ${\sum }_{n=1}^{+\infty }{f}_n$ is absolutely and uniformly convergent in A.

Proof. Let ${\sum }_{n=1}^{+\infty }{M}_n$ be a convergent series of real numbers and $\left(f_n\right)$ be a sequence of functions with common domain A such that $\mid f_n\left(x\right)\mid \le M_n$, for all $x\in A$ and all $n\in \mathbb{N}$.

Let x be an element of A. Since $\mid f_n\left(x\right)\mid \le M_n$, for each $n\in \mathbb{N}$, then the series ${\sum }_{n=1}^{+\infty }\mid f_n\left(x\right)\mid$ is convergent in A. Thus, ${\sum }_{n=1}^{+\infty }{f}_n\left(x\right)$ is absolutely convergent for any $x\in A$.

Regarding uniform convergence, since ${\sum }_{n=1}^{+\infty }{M}_n$ is convergent, then the sequence of partial sums

$$S_n=\sum _{k=1}^n{M}_k$$

is also convergent and, therefore, it is a Cauchy sequence. Then, for all $\epsilon >0$, there exists a $m\in \mathbb{N}$ such that, for all $n>m$ and for all $k\in \mathbb{N}$, we have

$$M_{n+1}+M_{n+2}+\cdots +M_{n+k}<\epsilon .$$

We can, therefore, write:

$$\begin{array}{ll}& \mid f_{n+1}\left(x\right)+f_{n+2}\left(x\right)+\cdots +f_{n+k}\left(x\right)\mid \\ \le & \mid f_{n+1}\left(x\right)\mid +\mid f_{n+2}\left(x\right)\mid +\cdots +\mid f_{n+k}\left(x\right)\mid \\ \le & M_{n+1}+M_{n+2}+\cdots +M_{n+k}\\ <& \epsilon ,\end{array}$$

for all $x\in A$. Hence, by Cauchy’s uniform convergence criterion for series of functions, we can conclude that the series ${\sum }_{n=1}^{+\infty }{f}_n$ is uniformly convergent. □

Example 3.36. In this example, we will show that the series of functions defined as

$$\sum _{n=1}^{+\infty }\frac{cos\left(\mathit{nx}\right)}{n^3}$$

converges uniformly in $\mathbb{R}$.

Indeed, we have that

$$\left|\frac{cos\left(\mathit{nx}\right)}{n^3}\right|\le \frac{1}{n^3},$$

$\forall x\in \mathbb{N}$.

Now, since ${\sum }_{n=1}^{+\infty }\frac{1}{n^3}$ is convergent, then, by the Weierstrass M-Test, we can conclude that the series ${\sum }_{n=1}^{+\infty }\frac{cos\left(\mathit{nx}\right)}{n^3}$ is uniformly convergent in $\mathbb{R}$.

We finish this subsection by presenting three results that we consider to be important to refer to in the context of this survey on function series.

Theorem 3.24. Let $\left(f_n\right)$ be a sequence of continuous functions in the domain $\left[a,b\right]\subset \mathbb{R}$. If ${\sum }_{n=1}^{+\infty }{f}_n$ is uniformly convergent to a function f in [a, b], then f is continuous in [a, b].

Theorem 3.25. Let $\left(f_n\right)$ be a sequence of integrable functions in the domain $\left[a,b\right]\subset \mathbb{R}$. If ${\sum }_{n=1}^{+\infty }{f}_n$ is uniformly convergent to a function f in [a, b], then f is integrable in [a, b] and we have that

$${\int }_a^{b}f\left(x\right)\mathit{dx}=\sum _{n=1}^{+\infty }{\int }_a^b{f}_n\left(x\right)\mathit{dx}.$$

Theorem 3.26. Let $\left(f_n\right)$ be a sequence of class $C^1$ functions in the domain $\left[a,b\right]\subset \mathbb{R}$ such that:

1. there is a $c\in \left[a,b\right]$ such that the series ${\sum }_{n=1}^{+\infty }{f}_n\left(c\right)$ is convergent and

2. the series ${\sum }_{n=1}^{+\infty }{f}_n^{\prime }$ is uniformly convergent to a function g in [a, b].

Then, the function series ${\sum }_{n=1}^{+\infty }{f}_n$ is uniformly convergent to a differentiable function f in [a, b] such that $f^{\prime }=g$ on [a, b] and

$$f^{\prime }\left(x\right)=\sum _{n=1}^{+\infty }{f}_n^{\prime }\left(x\right),\forall x\in \left[a,b\right].$$

3.5 Power series

In this section, we will study a particular class of function series that hold a great deal of importance for their applications, specifically regarding function analysis.

3.5.1 Definition and properties

In a very simple way, we can define a power series as an infinite sum like the following:

$$\sum _{n=0}^{+\infty }{a}_n{x}^n=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n+\cdots ,$$

where each $a_i,i\in \mathbb{N}$, is a real number.

The type of power series presented above is normally referred to as a power series centered at 0. A more general definition of a power series centered at any real number $x_0$ is given next.

Definition 3.14 (Power series). A power series centered at $x_0$ is a function series ${\sum }_{n=0}^{+\infty }{f}_n$, where each function $f_n$ can be defined as

$$f_n\left(x\right)=a_n{\left(x-x_0\right)}^n,$$

with $a_n\in \mathbb{R},\forall n\in {\mathbb{N}}_0$.

Given a power series centered at $x_0$, it is easy to observe that $x_0$ belongs to its convergence domain. However, what can we conclude on the convergence of a power series that we know that is convergent at a certain point $x=b?$ The following result sheds some light on this matter.

Theorem 3.27. Let ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ be a power series centered at 0, with $a_n\in \mathbb{R},\forall n\in \mathbb{N}$. Then:

1. if ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ converges at a non-zero value c, then ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ converges absolutely at any value x such that $\mid x\mid <\mid c\mid ;$

2. if ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ diverges at a certain non-zero value c, then ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ diverges for any x such that $\mid x\mid >\mid c\mid$.

Proof. Let ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ be a power series centered at 0, with $a_n\in \mathbb{R},\forall n\in \mathbb{N}$. Let $c\in \mathbb{R}\backslash \left\{0\right\}$.

1. If the series ${\sum }_{n=0}^{+\infty }{a}_n{c}^n$ is convergent, then we have that

   $$\underset{n\to +\infty }{lim}{a}_n{c}^n=0.$$

   But since the sequence whose general term is $a_n{c}^n$ is convergent, then it is bounded. Therefore, there exists a real positive number L such that

   $$\mid a_n{c}^n\mid <L,\forall n\in \mathbb{N}.$$

   Now, let $b\in \mathbb{R}$ such that $\mid b\mid <\mid c\mid$ and denote by $S_n\left(b\right)$ the sequence of partial sums of the series ${\sum }_{n=0}^{+\infty }\mid a_n{b}^n\mid .$ Then, we have:

   $$\begin{array}{lll}{S}_n\left(b\right)& =& \sum _{k=0}^n\mid a_k{b}^k\mid \\ & =& \sum _{k=0}^n\mid a_k\mid {\left|b\right|}^k\\ & =& \sum _{k=0}^n\mid a_k\mid \mid c^k\mid \frac{{\left|b\right|}^k}{{\left|c\right|}^k}\\ & =& \sum _{k=0}^n\mid a_k{c}^k\mid {\left|\frac{b}{c}\right|}^k\\ & \le & \sum _{k=0}^{n}L{\left|\frac{b}{c}\right|}^k\\ & =& L\sum _{k=0}^n{\left|\frac{b}{c}\right|}^k\end{array}$$

   Hence, we have

   $$S_n\left(b\right)\le L\left|\frac{b}{c}\right|\frac{1-{\left|\frac{b}{c}\right|}^n}{1-\left|\frac{b}{c}\right|}$$

   and, consequently,

   $$S_n\left(b\right)\le L\left|\frac{b}{c}\right|\frac{1}{1-\left|\frac{b}{c}\right|}.$$

   Now, since $S_n\left(b\right)$ is an increasing sequence of positive terms that is bounded from above, then we conclude that $S_n\left(b\right)$ is convergent and, therefore, the series ${\sum }_{k=0}^{+\infty }\mid a_k{b}^k\mid$ is convergent which means that the series ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ is absolutely convergent for any $x\in \mathbb{R},$ such that $\mid x\mid <\mid c\mid .$

2. If ${\sum }_{n=0}^{+\infty }{a}_n{c}^n$ is a divergent series, let $d\in \mathbb{R}$ be such that $\mid d\mid >\mid c\mid$.

   If ${\sum }_{n=0}^{+\infty }{a}_n{d}^n$ is convergent, then, by assertion (1.) proved before, the series ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ would be absolutely convergent on all values x such that $\mid x\mid <\mid d\mid$. But, since $\mid c\mid <\mid d\mid ,$ this would mean that the series ${\sum }_{n=0}^{+\infty }{a}_n{c}^n$ would be convergent, which is absurd because we are supposing precisely the contrary. Therefore, we conclude that ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ is divergent for any $x\in \mathbb{R}$ such that $\mid x\mid >\mid c\mid$.

□

The assertions of Theorem 3.27 allow us to conclude that:

• if the power series ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ is convergent at $c\in \mathbb{R}\backslash \left\{0\right\}$, then it is also absolutely convergent in the interval $]-\mid c\mid ,\mid c\mid [$ and

• if the power series ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ diverges at a point $c\in \mathbb{R}\backslash \left\{0\right\}$, then it is divergent in the union of intervals $]-\infty ,-\mid c\mid \left[\cup \right]\mid c\mid ,+\infty [$.

If there exists a real positive number R such that the power series ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ converges absolutely in the interval ] − R, R[ and is divergent in the interval $]-\infty ,-R\left[\cup \right]R,+\infty [$ (where at the points –R and R the convergence of the series needs to be checked), then this interval ] − R, R[ is called the interval of convergence while the number R is called the radius of convergence both regarding the power series ${\sum }_{n=0}^{+\infty }{a}_n{x}^n$ in x. If there is not such a real number, then we define the radius of convergence, R, as $+\infty$.

Note that, although we studied the particular case of a power series centered at 0, analogous conclusions can be drawn in the general case of a power series centered at any given point $x_0\in \mathbb{R}.$

Next we will show how we can obtain the radius of convergence of power series.

Theorem 3.28. Let $\left(a_n\right)$ be a sequence of real numbers such that:

$$\exists m\in \mathbb{N}:\forall n\in \mathbb{N},n>m\Rightarrow a_n\ne 0$$

and ${\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ be a power series centered at $x_0\in \mathbb{R}$. If $lim\left|\frac{a_n}{a_{n+1}}\right|$ exists, then the radius of convergence of this power series, R, is given by

$$R=lim\left|\frac{a_n}{a_{n+1}}\right|.$$

Proof. Let ${\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ be a power series satisfying the conditions of Theorem 3.28 and suppose that $lim\left|\frac{a_n}{a_{n+1}}\right|$ exists.

We will divide this proof in three cases.

1. If $lim\left|\frac{a_{n+1}}{a_n}\right|=0^{+}$, then the power series ${\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ is absolutely convergent for any x. Indeed, in this case we have:

   $$\begin{array}{lll}lim\frac{\left|a_{n+1}{\left(x-x_0\right)}^{n+1}\right|}{\left|a_n{\left(x-x_0\right)}^n\right|}& =& lim\left(\left|\frac{a_{n+1}}{a_n}\right|\frac{{\left|x-x_0\right|}^{n+1}}{{\left|x-x_0\right|}^n}\right)\\ & =& lim\left(\left|\frac{a_{n+1}}{a_n}\right|\left|x-x_0\right|\right)\\ & =& 0\left|x-x_0\right|\\ & =& 0<1.\end{array}$$

   Therefore, by the ratio test, the series ${\sum }_{n=1}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ is absolutely convergent for any $x\in \mathbb{R}$ and, consequently, the radius of convergence of the power series is $R=+\infty .$ Observe that

   $$lim\left|\frac{a_n}{a_{n+1}}\right|=lim\frac{1}{\left|\frac{a_{n+1}}{a_n}\right|}=\frac{1}{0^{+}}=+\infty .$$

   Hence, $R=lim\left|\frac{a_n}{a_{n+1}}\right|$.

2. If $lim\left|\frac{a_{n+1}}{a_n}\right|=+\infty$, then we have:

   $$\begin{array}{lll}lim\frac{\left|a_{n+1}{\left(x-x_0\right)}^{n+1}\right|}{\left|a_n{\left(x-x_0\right)}^n\right|}& =& lim\left(\left|\frac{a_{n+1}}{a_n}\right|\frac{{\left|x-x_0\right|}^{n+1}}{{\left|x-x_0\right|}^n}\right)\\ & =& lim\left(\left|\frac{a_{n+1}}{a_n}\right|\left|x-x_0\right|\right)\\ & =& +\infty \times \left|x-x_0\right|\\ & =& +\infty ,\end{array}$$

   if $x\ne x_0$. If $x=x_0$, then

   $$lim\frac{\left|a_{n+1}{x}^{n+1}\right|}{\left|a_n{x}^n\right|}=0.$$

   Therefore, the power series converges only at $x=x_0$. Hence,

   $$R=lim\left|\frac{a_n}{a_{n+1}}\right|=\frac{1}{+\infty }=0.$$

3. If $lim\left|\frac{a_{n+1}}{a_n}\right|=L\in \mathbb{R}$, then we have:

$$\begin{array}{lll}lim\frac{\left|a_{n+1}{\left(x-x_0\right)}^{n+1}\right|}{\left|a_n{\left(x-x_0\right)}^n\right|}& =& lim\left(\left|\frac{a_{n+1}}{a_n}\right|\frac{{\left|x-x_0\right|}^{n+1}}{{\left|x-x_0\right|}^n}\right)\\ & =& lim\left|\frac{a_{n+1}}{a_n}\right|\left|x-x_0\right|\\ & =& L\left|x-x_0\right|.\end{array}$$

Therefore, if

$$L\mid x-x_0\mid <1\iff \mid x-x_0\mid <\frac{1}{L},$$

then the power series ${\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ is absolutely convergent.

Hence, the radius of convergence of the power series is

$$R=\frac{1}{L}=\underset{n\to +\infty }{lim}\left|\frac{a_n}{a_{n+1}}\right|.$$

□

If R is the radius of convergence of a power series ${\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ in x, then the interval of convergence of the power series is the interval $]x_0-R,x_0+R[$. Note that, in general, the interval of convergence and the domain of convergence of the power series are not the same, as the latter may include the extreme points of the former in the case that the power series converges in such points.

Example 3.37. In this example, we will find the domain of convergence of the power series defined as:

$$\sum _{n=0}^{+\infty }{5}^n{x}^n.$$

Applying Theorem 3.28, we have:

$$\begin{array}{lll}R& =& lim\frac{\left|a_n\right|}{\left|a_{n+1}\right|}\\ & =& lim\frac{5^n}{5^{n+1}}\\ & =& lim\frac{1}{5}\\ & =& \frac{1}{5}.\end{array}$$

Hence, the interval of convergence of the given power series is: $\left]-\frac{1}{5},\frac{1}{5}\right[$.

Now let’s analyze the power series for $x=\frac{1}{5}$ and $x=-\frac{1}{5}$.

If $x=\frac{1}{5},$ the power series becomes the series ${\sum }_{n=0}^{\infty }1$ which is divergent since $lim{a}_n=lim1=1\ne 0$.

If $x=-\frac{1}{5}$, the power series becomes the series ${\sum }_{n=0}^{\infty }{\left(-1\right)}^n$ which is also divergent since $lim{a}_n=lim{\left(-1\right)}^n$ does not exist.

Hence, the domain of convergence of the power series ${\sum }_{n=0}^{+\infty }{5}^n{x}^n$ is the interval $\left]-\frac{1}{5},\frac{1}{5}\right[,$ that is, in this case, the domain of convergence corresponds to the interval of convergence.

We can also compute the radius of convergence through the root test presented before.

Theorem 3.29. Let $\left(a_n\right)$ be a sequence of real numbers such that:

$$\exists m\in \mathbb{N}:\forall n\in \mathbb{N},n>m\Rightarrow a_n\ne 0$$

and ${\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ be a power series centered at $x_0\in \mathbb{R}$. If $lim\frac{1}{\sqrt[n]{\mid a_n\mid }}$ exists, then the radius of convergence of this power series, R, is given by

$$R=lim\frac{1}{\sqrt[n]{\mid a_n\mid }}.$$

Proof. The proof is similar to the one given for Theorem 3.28. We leave it as an exercise for the reader. □

Example 3.38. In this example, we will determine the domain of convergence of the power series defined by

$$\sum _{n=0}^{+\infty }\frac{{\left(x-1\right)}^n}{n^n}.$$

By Theorem 3.29, we have:

$$\begin{array}{lll}R& =& lim\frac{1}{\sqrt[n]{\mid a_n\mid }}\\ & =& lim\frac{1}{\sqrt[n]{\frac{1}{n^n}}}\\ & =& lim\frac{1}{\frac{1}{n}}\\ & =& \frac{1}{0^{+}}\\ & =& +\infty .\end{array}$$

Therefore, the domain of convergence of the given power series is the interval of convergence $]-\infty ,+\infty [$ or simply the set $\mathbb{R}$.

The following two results relate power series convergence with the concepts of differentiation and integration. The proofs are left out because they require some mathematical tools outside the scope of this book.

Theorem 3.30. Let ${\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ be a power series whose radius of convergence is $R>0,$ convergent to the function f in $]x_0-R,x_0+R[$. Then, the power series ${\sum }_{n=1}^{+\infty }{\mathit{na}}_n{\left(x-x_0\right)}^{n-1}$ has radius of convergence equal to R and converges to the derivative of f, $f^{\prime }$, in $]x_0-R,x_0+R[.$ Additionally, this series converges uniformly in any closed interval $\left[a,b\right]\subset ]x_0-R,x_0+R[$.

A couple of examples are presented next.

Example 3.39. The series ${\sum }_{n=0}^{+\infty }{x}^n$ converge to the function $f\left(x\right)=\frac{1}{1-x}$ and the radius of this series is $R=1$. Hence, the series ${\sum }_{n=1}^{+\infty }{\mathit{nx}}^{n-1}$ is convergent to the function $f^{\prime }\left(x\right)=\frac{1}{{\left(1-x\right)}^2}$ in $]-1,1[$. Therefore, we can write

$$f^{\prime }\left(x\right)=\sum _{n=1}^{+\infty }{\mathit{nx}}^{n-1},\forall x\in ]-1,1[.$$

As for the extreme points of the interval $]-1,1[$, it is easy to conclude that the series ${\sum }_{n=1}^{+\infty }{\mathit{nx}}^{n-1}$ is divergent if $x=1$ or $x=-1$.

Example 3.40. In this example, we will consider the power series defined by:

$$\sum _{n=2}^{+\infty }\frac{x^n}{n\left(n-1\right)}.$$

We have that the radius of convergence of this series is $R=1$ and the domain of convergence is [–1, 1].

However, the series

$$\sum _{n=2}^{+\infty }\frac{{\mathit{nx}}^{n-1}}{n\left(n-1\right)}=\sum _{n=2}^{+\infty }\frac{x^{n-1}}{\left(n-1\right)}$$

with radius of convergence 1 is convergent at $x=-1$ but diverges at $x=1$. Thus, its domain of convergence is $[-1,1[$.

Theorem 3.31. Let ${\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ be a power series whose radius of convergence is $R>0,$ convergent to the function f in $]x_0-R,x_0+R[.$ Then, the power series ${\sum }_{n=0}^{+\infty }{a}_n\frac{x^{n+1}}{n+1}$ has radius of convergence equal to R and converges to the primitive of f, such that its value at $x_0$ is 0, in $]x_0-R,x_0+R[.$ Additionally, this series is uniformly convergent in any closed interval $\left[a,b\right]\subset ]x_0-R,x_0+R[$ and we have

$$\int f\left(x\right)\mathit{dx}=\sum _{n=1}^{+\infty }\frac{a_n}{n+1}{\left(x-x_0\right)}^{n+1},\forall x\in ]x_0-R,x_0+R[.$$

3.5.2 Representation of a function as a power series

Arguably, the most notable application of power series is the representation of functions that are infinitely differentiable at a real number of its domain. By representing a function by its power series, we can easily obtain approximations of that function at a given point with an associated error as small as we want. It was, therefore, natural not to finish this section on power series without summarizing the key aspects on this important application.

Indeed, in some situations, given a function f, it is important to obtain a power series that converges to f at least at a certain domain. Such a power series is said to represent the function f in the domain of convergence.

Before we proceed any further, let’s establish that this representation of a function by a power series is unique. In fact, if we have simultaneously that $f\left(x\right)={\sum }_{n=0}^{+\infty }{a}_n{x}^n$ and $f\left(x\right)={\sum }_{n=0}^{+\infty }{b}_n{x}^n$ for all x in a certain domain D, then we have

$$\begin{array}{ll}& \sum _{n=0}^{+\infty }{a}_n{x}^n=\sum _{n=0}^{+\infty }{b}_n{x}^n\\ \iff & \sum _{n=0}^{+\infty }\left(a_n-b_n\right)x^n=0\end{array}$$E30

$$\iff a_0-b_0+\left(a_1-b_1\right)x+\left(a_2-b_2\right)x^2+\cdots$$E31

Since 0 naturally belongs to the domain of convergence D, if we make $x=0$ in (31), we obtain:

$$a_0-b_0=0\iff a_0=b_0.$$

Now, differentiating equality (30) established above, with respect to x, we obtain

$$\begin{array}{c}\begin{array}{ll}& \sum _{n=1}^{+\infty }n\left(a_n-b_n\right)x^{n-1}=0\\ \iff & a_1-b_1+2\left(a_2-b_2\right)x+3\left(a_3-b_3\right)x^2+\cdots \end{array}\end{array}$$E32

Again, since $0\in D$, substituting $x=0$ in (32) we obtain

$$a_1-b_1=0\iff a_1=b_1.$$

Applying this reasoning successively, we can conclude that $a_n=b_n,\forall n\in \mathbb{N}$, that is, the two developments of f as a power series are equal. A similar conclusion would be drawn if the series were centered in a given non-zero point.

It is known that, if f admits derivatives until the order $n+1$ in an open interval $I=]a,b[$, with $a<b$, centered at $x=x_0$, then f can be written in the following way:

$$\begin{array}{lll}f\left(x\right)& =& f\left(x_0\right)+\frac{f^{\prime }\left(x_0\right)}{1!}\left(x-x_0\right)+\frac{{f^{\prime }}^{\prime }\left(x_0\right)}{2!}{\left(x-x_0\right)}^2+\frac{{{f^{\prime }}^{\prime }}^{\prime }\left(x_0\right)}{3!}{\left(x-x_0\right)}^3\\ & +& \cdots +\frac{f^{\left(n\right)}\left(x_0\right)}{n!}{\left(x-x_0\right)}^n+r_n\left(x_0,x\right)\\ & =& \sum _{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right)}{k!}{\left(x-x_0\right)}^k+r_n\left(x_0,x\right),\end{array}$$E33

for $x\in I$, where $r_n\left(x_0,x\right)$ is a remainder of this development of f given by

$$r_n\left(x_0,x\right)=\frac{f^{\left(n+1\right)}\left(c\right)}{\left(n+1\right)!}{\left(x-x_0\right)}^{n+1},$$

for some $c\in ]x_0,x[$ if $x>x_0$, or $c\in ]x,x_0[$ if $x<x_0$.

In the cases that we have that

$$\underset{n\to +\infty }{lim}{r}_n\left(a,x\right)=0,$$E34

then, applying limits to equality (33), we deduce

$$\begin{array}{ll}& \underset{n\to +\infty }{lim}f\left(x\right)=\underset{n\to +\infty }{lim}\sum _{k=0}^n\frac{f^{\left(k\right)}\left(a\right)}{k!}{\left(x-a\right)}^k+\underset{n\to +\infty }{lim}{r}_n\left(a,x\right)\\ \iff & f\left(x\right)=\sum _{n=0}^{+\infty }\frac{f^{\left(n\right)}\left(x_0\right)}{n!}{\left(x-x_0\right)}^n.\end{array}$$E35

The development of the function f presented in (35) is called the Taylor series of the function f at the point $x=x_0$. In the particular case where the series is centered at the point $x_0=0$, it is called a Maclaurin series instead of a Taylor series centered at 0. The names of these series were given after the British mathematicians who developed and applied them in their respective works: the English mathematician Brook Taylor (1685–1731) and the Scottish mathematician Colin Maclaurin (1698–1746).

From what was presented above, it should be clear that the Taylor series represents the function f if and only if equality (34) holds. However, by reading this very brief survey on the Taylor series, one might assume that it is enough for the function f to be infinitely differentiable to be able to be represented by a Taylor series. In general, this is not true, that is, although we can always construct the Taylor series of f if f is infinitely differentiable, the convergence

$$f\left(x\right)=\sum _{n=0}^{+\infty }\frac{f^{\left(n\right)}\left(x_0\right)}{n!}{\left(x-x_0\right)}^n$$

is not guaranteed. A classic example that enlightens this situation is the function f such that $f\left(0\right)=0$ and $f\left(x\right)=e^{-\frac{1}{x^2}}$, if $x\ne 0$. It is easy to see that

$$f\left(x\right)\ne \sum _{n=0}^{+\infty }\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n,$$

for all $x\ne 0$, since that in this case the Taylor series is the null series because all the derivatives of f at 0 are null.

The functions, f, that can be represented by a Taylor series at a certain domain D are, therefore, required to have a “very good behavior” that goes beyond being infinitely differentiable. These functions are usually called analytic which means simply that they can be written, for any point $x=x_0\in D$, as

$$f\left(x\right)=\sum _{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$$

for some coefficients $a_n$ and with a positive radius of convergence.

As we suggested in the beginning of this subsection, given a Taylor series at $x_0$ that represents a function f on a certain interval I containing $x_0$, we are in conditions to obtain an approximation of f at a point $x\in I$ by considering a finite partial sum of the Taylor series at $x_0$ up to the order that will provide the desired approximation significance. This partial sum up to the order, let’s say, m is called the Taylor polynomial of order m of f at the point $x_0$, which we denote by $T_m\left(x_0,x\right),$ and we can write, for $x\in I$:

$$\begin{array}{lll}f\left(x\right)& \approx & T_m\left(x_0,x\right)\\ & =& \sum _{k=0}^m\frac{f^{\left(k\right)}\left(x_0\right)}{k!}{\left(x-x_0\right)}^k\\ & =& f\left(x_0\right)+\frac{f^{\prime }\left(x_0\right)}{1!}{\left(x-x_0\right)}^1+\cdots +\frac{f^{\left(m\right)}\left(x_0\right)}{m!}{\left(x-x_0\right)}^m.\end{array}$$

Besides adding summands to the Taylor polynomial by increasing the value of m, we can also improve the quality of the approximation by considering a center of the power series, $x_0$, closer to the x in I on which we are evaluating the function f. By doing so, we make the successive powers ${\left(x-x_0\right)}^k$ smaller and, consequently, the Taylor polynomial will provide a more accurate approximation to f(x).

Example 3.41. In this example, we will illustrate the approximation technique presented before with the function $f\left(x\right)=e^x$.

For this function, we have the following Taylor series centered at 0:

$$\begin{array}{lll}f\left(x\right)& =& e^0+e^{0}x+\frac{e^0}{2!}{x}^2+\frac{e^0}{3!}{x}^2+\cdots \\ & =& 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots ,\end{array}$$

with the first Taylor polynomials of orders 1 to 3 being:

$$\begin{array}{lll}{T}_1\left(0,x\right)& =& 1+x;\\ T_2\left(0,x\right)& =& 1+x+\frac{x^2}{2};\\ T_3\left(0,x\right)& =& 1+x+\frac{x^2}{2}+\frac{x^3}{6}.\end{array}$$

In Figure 3, we show graphically how these approximations improve as they come closer to the function f with the increase of the order of the polynomial.

Figure 3.

Approximations with Taylor polynomials to the function $e^x$.

Let ${\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ be a power series with radius of convergence R. Using Theorem 3.30, we know that, if $f\left(x\right)={\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$, then the series ${\sum }_{n=1}^{+\infty }{\mathit{na}}_n{\left(x-x_0\right)}^{n-1}$ as also radius of convergence equal to R and

$$f^{\prime }\left(x\right)=\sum _{n=1}^{+\infty }{a}_n{\mathit{nx}}^{n-1},\forall x\in ]x_0-R,x_0+R[.$$

If we keep applying Theorem 3.30 recursively, we conclude that, for $x\in ]x_0-R,x_0+R[$, we have

$$\begin{array}{lll}{f^{\prime }}^{\prime }\left(x\right)& =& n\left(n-1\right)a_n{\left(x-b\right)}^{n-2};\\ {{f^{\prime }}^{\prime }}^{\prime }\left(x\right)& =& n\left(n-1\right)\left(n-2\right)a_n{\left(x-b\right)}^{n-3};\\ & ⋮& \\ f^{\left(p\right)}\left(x\right)& =& n\left(n-1\right)\cdots \left(n-p+1\right)a_n{\left(x-b\right)}^{n-p},\forall p\in \mathbb{N}.\end{array}$$

Therefore, a function $f\left(x\right)={\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$, for $x\in ]x_0-R,x_0+R[$, is infinitely differentiable at $x_0$ and its derivatives are such that $f^{\left(p\right)}\left(b\right)=p!a_p,\forall p\in \mathbb{N}$. We can, therefore, conclude that

$$a_p=\frac{f^{\left(p\right)}\left(x_0\right)}{p!},\forall p\in \mathbb{N}.$$

Hence, we conclude that if $f\left(x\right)={\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n,$ for $x\in ]x_0-R,x_0+R[$, then the development of f as a Taylor series at the point $x_0$ is precisely $f\left(x\right)={\sum }_{n=0}^{+\infty }{a}_n{\left(x-x_0\right)}^n$ and this development is valid in $]x_0-R,x_0+R[$, while for $x<x_0-R$ or $x>x_0+R$ the Taylor series is divergent. The points $x=x_0-R$ and $x=x_0+R$ must be objective of a particular analysis.

Next, we present an example where we deduce the Maclaurin series of a function.

Example 3.42. Consider the function f defined by

$$f\left(x\right)=\frac{1}{{\left(1-x\right)}^2}$$

for any x belonging to the interval $]-\infty ,1[$. Let’s determine the Maclaurin series of f and its domain of convergence.

We know that

$$\frac{1}{1-x}=1+x+x^2+\cdots +=\sum _{n=0}^{+\infty }{x}^n,$$

for $x\in ]-1,1[$. The radius of convergence of this series is

$$R=lim\sqrt[n]{1}=1.$$

Now, applying Theorem 3.30, we have that

$$\begin{array}{ll}& {\left(\frac{1}{1-x}\right)}^{\prime }=\sum _{n=1}^{+\infty }{\mathit{nx}}^{n-1}\\ \iff & \frac{1}{{\left(1-x\right)}^2}=\sum _{n=1}^{+\infty }{\mathit{nx}}^{n-1},\end{array}$$

for $x\in ]-1,1[$.

Therefore, the Maclaurin series of f is ${\sum }_{n=1}^{+\infty }{\mathit{nx}}^{n-1}$ and this development is valid for $x\in ]-1,1[.$ Additionally, since the series ${\sum }_{n=1}^{+\infty }{\mathit{nx}}^{n-1}$ is divergent at $x=1$ and $x=-1$, then the function f can be developed as a Maclaurin series only in the interval $]-1,1[$.

We finish this section with a list of the most commonly used Maclaurin series presented in Table 2.

f(x)	Maclaurin series	Conv. domain
$e^x$	$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots =\sum _{n=0}^{+\infty }\frac{x^n}{n!}$	$\mathbb{R}$
ln(1 + x)	$x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots =\sum _{n=1}^{+\infty }{\left(-1\right)}^{n+1}\frac{x^n}{n}$	] − 1, 1]
ln(1 − x)	$-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots =\sum _{n=1}^{+\infty }\frac{-x^n}{n}$	[−1, 1[
$\frac{1}{1-x}$	$1+x+x^2+x^3+\cdots =\sum _{n=0}^{+\infty }{x}^n$	] − 1, 1[
$\frac{1}{{\left(1-x\right)}^2}$	$1+2x+3x^2+\cdots =\sum _{n=1}^{+\infty }{\mathit{nx}}^{n-1}$	] − 1, 1[
${\left(1+x\right)}^n$	$1+\mathit{nx}+\frac{n\left(n-1\right)}{2!}+\cdots =\sum _{k=0}^{+\infty }\left(\begin{array}{c}n\\ k\end{array}\right)x^k$	] − 1, 1[
sin(x)	$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots =\sum _{n=0}^{+\infty }\frac{{\left(-1\right)}^n}{\left(2n+1\right)!}{x}^{2n+1}$	$\mathbb{R}$
cos(x)	$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots =\sum _{n=0}^{+\infty }\frac{{\left(-1\right)}^n}{\left(2n\right)!}{x}^{2n}$	$\mathbb{R}$
tan(x)	$x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\cdots$	$\left]-\frac{\pi }{2},\frac{\pi }{2}\right[$
arcsin(x)	$x+\frac{x^3}{6}+\frac{3x^5}{40}+\cdots =\sum _{n=0}^{+\infty }\frac{\left(2n\right)!}{4^n{\left(n!\right)}^2\left(2n+1\right)}{x}^{2n+1}$	[−1, 1]
arccos(x)	$\frac{\pi }{2}-x-\frac{x^3}{6}-\frac{3x^5}{40}-\cdots =\frac{\pi }{2}\sum _{n=0}^{+\infty }\frac{\left(2n\right)!}{4^n{\left(n!\right)}^2\left(2n+1\right)}{x}^{2n+1}$	[−1, 1]
arctan(x)	$x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots =\sum _{n=0}^{+\infty }\frac{{\left(-1\right)}^n}{2n+1}{x}^{2n+1}$	[−1, 1]
sinh(x)	$x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots =\sum _{n=0}^{+\infty }\frac{1}{\left(2n+1\right)!}{x}^{2n+1}$	$\mathbb{R}$
cosh(x)	$1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots =\sum _{n=0}^{+\infty }\frac{1}{\left(2n\right)!}{x}^{2n}$	$\mathbb{R}$
tanh(x)	$x-\frac{x^3}{3}+\frac{2x^5}{15}-\frac{17x^7}{315}+\cdots$	$\left]-\frac{\pi }{2},\frac{\pi }{2}\right[$

Table 2.

Most common Maclaurin series.

3.6 Suggested exercises

Convergence of series

1. Classify, using the definition of convergent series, the following series.

   1. ${\sum }_{n=1}^{+\infty }{\left(\frac{1}{5}\right)}^n;$

   2. ${\sum }_{n=1}^{+\infty }\left(\sqrt{n+2}-\sqrt{n+1}\right);$

   3. ${\sum }_{n=1}^{+\infty }\left(\frac{1}{n^2}-\frac{1}{{\left(n+1\right)}^2}\right);$

   4. ${\sum }_{n=2}^{+\infty }ln\left(\frac{n+1}{n}\right);$

   5. ${\sum }_{n=1}^{+\infty }\left(\frac{1}{\left(n+2\right)!}-\frac{1}{n!}\right).$

2. Classify the following series and, if possible, obtain their sum.

   1. ${\sum }_{n=1}^{+\infty }\frac{1}{3^n};$

   2. ${\sum }_{n=1}^{+\infty }\frac{1}{n\left(n+1\right)};$

   3. ${\sum }_{n=1}^{+\infty }\left(\frac{1}{2^n}+\frac{1}{\left(n+1\right)\left(n+2\right)}\right);$

   4. ${\sum }_{n=1}^{+\infty }\left(\frac{1}{n\left(n+3\right)}\right);$

   5. ${\sum }_{n=1}^{+\infty }\left(\frac{n+1}{n^2{\left(n+2\right)}^2}\right);$

   6. ${\sum }_{n=1}^{+\infty }\frac{3}{2^n};$

   7. ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^{n+1}\frac{3^{n+1}}{5^n};$

   8. ${\sum }_{n=1}^{+\infty }\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)};$

   9. ${\sum }_{n=1}^{+\infty }\frac{3^n+4^n+5^n}{5^n}.$

3. Show that

   1. ${\sum }_{n=1}^{+\infty }\frac{1}{\left(2n-1\right)\left(2n+5\right)}=\frac{23}{90};$

   2. ${\sum }_{n=1}^{+\infty }\left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^2+n}}\right)=1;$

   3. ${\sum }_{n=1}^{+\infty }ln\left(\frac{n^2+2n}{{\left(n+1\right)}^2}\right)=ln\left(\frac{1}{2}\right).$

4. Obtain the general term of the series ${\sum }_{n=1}^{+\infty }{u}_n$ knowing that

   $$S_n=\frac{n}{n+1}.$$

5. Find the general term of the following series (suppose that the terms follow the same rule of formation) and analyze their convergence. Also, in each case where convergence occurs, calculate the series sum.

   1. $2+4+8+16+32+\cdots ;$

   2. $2-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\cdots ;$

   3. $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\cdots ;$

   4. $1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots .$

   Series of non-negative terms

6. Prove Theorem 3.8 using the Integral Test.

7. Use the comparison test to analyze the convergence of the following series.

   1. ${\sum }_{n=1}^{+\infty }\frac{1}{n^2+n+1};$

   2. ${\sum }_{n=1}^{+\infty }\frac{1}{\sqrt{n^3+1}};$

   3. ${\sum }_{n=2}^{+\infty }\frac{ln\left(n+1\right)}{n};$

   4. ${\sum }_{n=1}^{+\infty }\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)};$

   5. ${\sum }_{n=1}^{+\infty }\frac{3^n+2^n}{7^n+5^n};$

   6. ${\sum }_{n=1}^{+\infty }\frac{1}{\sqrt{n+1}}.$

8. Show that the series

   $$\sum _{n=1}^{+\infty }\frac{ln\left(3\right)ln\left(4\right)ln\left(5\right)\cdots ln\left(n\right)}{\left(n+2\right)!}$$

   is convergent.

9. Use the limit comparison test to analyze the convergence of the following series of real numbers.

   1. ${\sum }_{n=1}^{+\infty }\frac{n^2+1}{n^4+n^2+1};$

   2. ${\sum }_{n=1}^{+\infty }\frac{n+1}{n\sqrt{n}+n+1};$

   3. ${\sum }_{n=1}^{+\infty }\frac{3^n+4^n}{7^n+6^n};$

   4. ${\sum }_{n=1}^{+\infty }\frac{1}{\sqrt[3]{n^2+n+1}};$

   5. ${\sum }_{n=1}^{+\infty }\frac{n}{2n^2+1};$

   6. ${\sum }_{n=1}^{+\infty }\frac{\sqrt[3]{n}}{\left(n+2\right)\sqrt{n}};$

   7. ${\sum }_{n=1}^{+\infty }\frac{1}{{\left({\left(-1\right)}^n+5\right)}^{5n}};$

   8. ${\sum }_{n=1}^{+\infty }\frac{1}{n\left(n+1\right)\left(n+2\right)(n+3}.$

10. Analyze the convergence of the following series by applying the ratio test.

    1. ${\sum }_{n=1}^{+\infty }\frac{1}{n!};$

    2. ${\sum }_{n=1}^{+\infty }\frac{n}{5^n};$

    3. ${\sum }_{n=1}^{+\infty }\frac{\sqrt{n+1}}{6^n};$

    4. ${\sum }_{n=1}^{+\infty }\frac{5^n}{n^2};$

    5. ${\sum }_{n=2}^{+\infty }\frac{n^n}{n!};$

    6. $\frac{n^4}{{\left(ln5\right)}^n};$

    7. ${\sum }_{n=1}^{+\infty }\frac{n^4\cdot 5^n}{n!};$

    8. ${\sum }_{n=1}^{+\infty }\frac{5^n}{7^n+8^n};$

    9. ${\sum }_{n=1}^{+\infty }\frac{n+3}{5^n{\left(n+1\right)}^2}.$

11. Use the root test to analyze the convergence of the following series.

    1. ${\sum }_{n=1}^{+\infty }\frac{n^n}{5^{2n+1}};$

    2. ${\sum }_{n=1}^{+\infty }{n}^n;$

    3. ${\sum }_{n=1}^{+\infty }{\left(\frac{n}{n+1}\right)}^{n^2};$

    4. ${\sum }_{n=1}^{+\infty }{\left(\frac{n^2}{2n^2+1}\right)}^{2n};$

    5. ${\sum }_{n=1}^{+\infty }{\left(\frac{n+1}{n}\right)}^n;$

    6. ${\sum }_{n=1}^{+\infty }\frac{1}{{\left(lnn\right)}^n};$

    7. ${\sum }_{n=1}^{+\infty }\frac{1}{{\left(n+1\right)}^n}.$

12. Analyze the convergence of the following series by applying the integral test.

    1. ${\sum }_3^{+\infty }\frac{1}{nlnn};$

    2. ${\sum }_{n=1}^{+\infty }\frac{n}{\left(4n-3\right)\left(4n-1\right)};$

    3. ${\sum }_{n=1}^{+\infty }\frac{1}{n^2+1};$

    4. ${\sum }_{n=1}^{+\infty }\frac{n}{1+n^2};$

    5. ${\sum }_{n=1}^{+\infty }\frac{1}{\sqrt{2n+1}};$

    6. ${\sum }_{n=1}^{+\infty }\frac{1}{n{ln}^{4}n};$

    7. ${\sum }_{n=1}^{+\infty }\frac{1}{\left(n+1\right)\sqrt{ln\left(n+1\right)}};$

    8. ${\sum }_{n=1}^{+\infty }{n}^3{e}^{-n^4};$

    9. ${\sum }_{n=1}^{+\infty }\left(n+1\right)e^{-\left(n+1\right)};$

13. Find the nature of convergence of the following series analyzing its general term.

    1. ${\sum }_{n=1}^{+\infty }\frac{n^2+2n+1}{n^2+2n+3};$

    2. ${\sum }_{n=1}^{+\infty }{\left(\frac{n+1}{n}\right)}^{2n};$

    3. ${\sum }_{n=1}^{+\infty }\frac{n}{\sqrt{2n^2+1}};$

    4. ${\sum }_{n=1}^{+\infty }\sqrt[n]{n};$

    5. ${\sum }_{n=1}^{+\infty }\frac{2^n}{n}.$

14. Show that, if the series of positive terms ${\sum }_{n=1}^{+\infty }{u}_n$ converges, then the series

    $$\sum _{n=1}^{+\infty }\left(\frac{n+1}{2n+1}{u}_n\right)$$

    also converges.

    Series of alternating terms

15. Show that the following series are convergent using the alternating series test and study their absolute convergence.

    1. ${\sum }_{n=1}^{+\infty }\frac{{\left(-1\right)}^n}{n^2};$

    2. ${\sum }_{n=1}^{+\infty }\frac{{\left(-1\right)}^{n-1}}{\sqrt[5]{n}};$

    3. ${\sum }_{n=1}^{+\infty }\frac{{\left(-1\right)}^n}{n{5}^n};$

    4. ${\sum }_{n=1}^{+\infty }\frac{{\left(-1\right)}^n}{n\left(n+1\right)\left(2n+1\right)};$

    5. ${\sum }_{n=1}^{+\infty }\frac{{\left(-1\right)}^n}{\left(3n\right)!};$

16. Study the convergence of each of the following series stating if the convergence is absolute or conditional.

    1. ${\sum }_{n=1}^{+\infty }sin\left(\mathit{n\pi }\right)\frac{2n+1}{n};$

    2. ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^{n}cos\left(\mathit{n\pi }\right);$

    3. ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^n\left(\frac{n+1}{3^n}\right);$

    4. ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^{n+2}\frac{2^n}{3^{n+1}};$

    5. ${\sum }_{N=1}^{+\infty }{\left(-1\right)}^n\frac{n^2}{n^4+n^2+1};$

    6. ${\sum }_{n=1}^{+\infty }\frac{sin\left(\mathit{n\pi }\right)}{n^4};$

    7. ${\sum }_{n=1}^{+\infty }cos\left(\mathit{n\pi }\right){\mathit{ne}}^n;$

    8. ${\sum }_{n=1}^{+\infty }\frac{cos\left(n\right)}{1-n^3}.$

    Function series

17. In each of the following cases, study if the given sequence of functions $\left(f_n\right)$ is pointwise or uniformly convergent on the given domain D.

    1. $f_n\left(x\right)=\frac{x^2}{n},D=]-\infty ,+\infty [;$

    2. $f_n\left(x\right)=\frac{1}{1+n^4{x}^2},D=\left[2,3\right];$

    3. $f_n\left(x\right)=\frac{x}{2+\mathit{nx}},D=[2,+\infty [;$

    4. $f_n\left(x\right)=\frac{sin\left(\mathit{nx}\right)}{n^4},D=\mathbb{R};$

    5. $f_n\left(x\right)=\sqrt{x^4+\frac{1}{n^2}},D=\mathbb{R};$

18. Show that the sequence of functions $\left(f_n\right)$, where $f_n\left(x\right)=x^n\left(1-x^n\right)$, is pointwise convergent but not uniformly convergent to the null function on the interval [0, 1].

19. Show that the sequence of functions $\left(f_n\right)$, where, for each $n\in \mathbb{N}$ we have $f_n:\left[0,1\right]\to \mathbb{R}$ and

    $$f_n\left(x\right)=\left\{\begin{array}{ll}\frac{n}{n-1}x,& \text{if}x\in \left[0,1-\frac{1}{n}\right]\\ -n\left(x-1\right),& \text{if}x\in ]1-\frac{1}{n},1]\end{array},\right.$$

    is pointwise convergent but not uniformly convergent to the function f such that $f\left(x\right)=x,\forall x\in [0,1[$ and $f\left(1\right)=0$.

20. Consider the sequence of functions $\left(f_n\right)$ such that

    $$\begin{array}{cccc}{f}_n:& \left[0,1\right]& \to & \mathbb{R}\\ & x& \mapsto & \left(1-x\right)x^n\end{array}.$$

    1. Show that, for all $x\in \left[0,1\right]$, the sequence of functions $\left(f_n\right)$ is pointwise convergent to the function f such that $f\left(x\right)=0$.

    2. Show that

       $$\underset{x\in \left[0,1\right]}{max}\left|f_n\left(x\right)-f\left(x\right)\right|={\left(\frac{n}{n+1}\right)}^n\frac{1}{n+1}.$$

    3. Show that the sequence $\left(f_n\right)$ is uniformly convergent to the function f presented in (a).

21. Consider the sequence of functions $\left(f_n\right)$ such that

    $$\begin{array}{cccc}{f}_n:& \left[0,1\right]& \to & \mathbb{R}\\ & x& \mapsto & \frac{n^2{x}^2}{1+n^4{x}^4}\end{array}.$$

    1. Show that $\left(f_n\right)$ is pointwise convergent to the null function f on [0, 1].

    2. Verify that $f_n\left(\frac{1}{n}\right)=\frac{1}{2}$ and justify why the sequence of functions $\left(f_n\right)$ does not converge uniformly to the null function f.

22. Using the Weierstrass M-test, show that the following series of functions are uniformly convergent on the domain presented in each item.

    1. ${\sum }_{n=1}^{+\infty }\frac{1}{n^4+x^8},D=\mathbb{R};$

    2. ${\sum }_{n=1}^{+\infty }\frac{sin\left(\mathit{nx}\right)}{n^4},D=\mathbb{R};$

    3. ${\sum }_{n=1}^{+\infty }{3}^{n}sin\left(\frac{x^n}{5^n}\right),D=\left[0,1\right];$

    4. ${\sum }_{n=1}^{+\infty }\frac{{\left(-1\right)}^n}{n{\left(x+3\right)}^3},D=[1,+\infty [.$

23. Consider the sequence of functions $\left(f_n\right)$ defined as $f_n:\left[0,1\right]\to \mathbb{R}$ such that

    $$f_n\left(x\right)=\left\{\begin{array}{ll}2n^{2}x,& \text{if}x\in \left[1,\frac{1}{2n}\right]\\ -2n\left(\mathit{nx}-1\right),& \text{if}x\in \left]\frac{1}{2n},\frac{1}{n}\right]\\ 0,& \text{if}x\in \left]\frac{1}{n},1\right]\end{array}\right..$$

    Show that $\left(f_n\right)$ is pointwise convergent but not uniformly convergent to the null function on [0, 1].

    [Suggestion: start by showing that the area between the null function and $f_n$ is equal to $\frac{1}{2}$.]

24. Show that the series

    $$\sum _{n=1}^{+\infty }\frac{2{\mathit{nx}}^2}{1+n^4{x}^4}$$

    is uniformly convergent on the interval [0, 1].

25. Show that if ${\sum }_{n=1}^{+\infty }{a}_n$ is absolutely convergent, then the series of functions ${\sum }_{n=1}^{+\infty }\frac{a_n{x}^{4n}}{1+x^{4n}}$ is uniformly convergent on the interval [0, 1].

    Power series

26. Find the radius of convergence and the interval of convergence of the following power series.

    1. ${\sum }_{n=1}^{+\infty }\frac{{\left(x-1\right)}^n}{n!};$

    2. ${\sum }_{n=1}^{+\infty }{5}^n{x}^n;$

    3. ${\sum }_{n=1}^{+\infty }\frac{x^n}{n};$

    4. ${\sum }_{n=1}^{+\infty }\frac{x^n}{n{2}^n};$

    5. ${\sum }_{n=1}^{+\infty }\frac{x^{2n}}{\sqrt{n}};$

    6. ${\sum }_{n=1}^{+\infty }\frac{x^n}{n^2}.$

27. Show that

    $$\frac{1}{{\left(1-x\right)}^2}=\sum _{n=0}^{+\infty }\left(n+1\right)x^n$$

    for $\mid x\mid <1$.

28. Show that

    $$\frac{6x}{5x^2-4x-1}=\sum _{n=1}^{+\infty }\left[{\left(-5\right)}^n-1\right]x^n$$

    for $\mid x\mid <\frac{1}{5}$.

29. Show that

    $$ln\left(1+x\right)=\sum _{n=1}^{+\infty }{\left(-1\right)}^{n-1}\frac{x^n}{n}$$

    for all $x\in ]-1,1]$.

30. Obtain the interval of convergence of the following power series.

    1. ${\sum }_{n=0}^{+\infty }\frac{{\left(x-3\right)}^n}{2^n};$

    2. ${\sum }_{n=0}^{+\infty }\frac{4^n}{\left(2n\right)!}{\left(x-1\right)}^n;$

    3. ${\sum }_{n=1}^{+\infty }\frac{1}{n}{x}^n;$

    4. ${\sum }_{n=1}^{+\infty }\frac{n}{2^n}{\left(x-2\right)}^n;$

    5. ${\sum }_{n=1}^{+\infty }\frac{n}{n+1}{x}^n.$

31. Find the domain of convergence of the following series.

    1. ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^n{x}^n;$

    2. ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^n\frac{x^n}{\sqrt{n}};$

    3. ${\sum }_{n=1}^{+\infty }\frac{x^n}{n!};$

    4. ${\sum }_{n=1}^{+\infty }{\left(-1\right)}^n\frac{n^2{x}^n}{3^n};$

    5. ${\sum }_{n=1}^{+\infty }\frac{n!{\left(x-1\right)}^n}{1\cdot 3\cdot 5\cdots \left(2n-1\right)!}.$

32. Consider the function $f\left(x\right)=e^{-x^2}$.

    1. Determine the Maclaurin series of f.

    2. Using the Taylor polynomial of order 4 of the series determined in (a), obtain an approximation of f(2).

33. Consider the function $g\left(x\right)=sin\left(2x-8\right)$.

    1. Determine the Taylor series of g at $x_0=4$.

    2. Using the Taylor polynomial of order 5 of the series determined in (a), obtain an approximation of g(5).

Solutions to suggested exercises

Sequences

1. 1.

   1. (a) $u_n=n^2-n$

   2. (b) $156=u_{13}$

   3. (c) $n\in \left\{\mathrm{21,22,23,24}\right\}$

2. 2.

   1. (a) $v_1=\frac{14}{5},v_2=\frac{5}{2},v_3=\frac{26}{11},v_4=\frac{16}{7},v_5=\frac{38}{17}$

   2. (b) No, it is not a term of $\left(v_n\right)$ since $n=\frac{14}{3}\notin \mathrm{\mathbb{N}}$

   3. (c) $v_{n+1}-v_n=\frac{-12}{\left(3n+5\right)\left(3n+2\right)}<0$

   4. (d) $2<v_n\le \frac{14}{5}$

   5. (e) $n=13$

3. 3.

   1. (a) $w_n=\frac{{\left(-1\right)}^n{n}^3}{{\left(n+1\right)}^2}$

   2. (b) $w_{10}=\frac{1000}{121}$

   3. (c) $lim{w}_n=+\infty$, for even n and $lim{w}_n=-\infty$, for odd n

4. 4.

   1. (a) $u_1=5,u_2=25,u_3=85,u_4=265$

   2. (c) $\left(u_n\right)$ is increasing $\left(u_{n+1}-u_n=\frac{10}{3}\times \left(3^{n+1}-3^n\right)>0\right)$

   3. (d) $lim{u}_n=+\infty$

5. 5.

   1. (a) $\left\{-1,-\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2},1\right\}$

   2. (b) $-1\le v_n\le 1$

   3. (c) $v_1<v_2>v_3$

6. 6.

   1. (a) $\left(w_n\right)$ is increasing

   2. (b) $-1\le w_n<5$

7. 7. 10

8. 8.

   1. (a) $v_5=-\frac{9}{13}$

   2. (b) Yes, it is the term $v_{13}$

   3. (c) $\left(v_n\right)$ is decreasing $\left(v_{n+1}-v_n=\frac{-8}{\left(2n+5\right)\left(2n+3\right)}<0\right)$

   4. (d) $-1<v_n\le -\frac{1}{5}$

9. 9. $\left(w_n\right)$ is increasing, since $w_{n+1}-w_n={\left(-1\right)}^{n+1}-{\left(-1\right)}^n+3>1>0$

10. 10.

    1. (a) $u_n=3n-1$

    2. (b) 63

    3. (c) 610

11. 11.

    1. (a) $v_{n+1}-v_n=-\frac{5}{3}$

    2. (b) $v_{n+1}-v_n<0$

    3. (c) –745

12. 12.

    1. (a) $d=4,\left(w_n\right)$ is increasing

    2. (b) $w_n=4n-8$

    3. (c) m = 15

13. 13.

    1. (a) $u_n=4\times 5^{1-n}$

    2. (b) $u_n$ is decreasing

    3. (c) $0<u_n\le 4$

    4. (d) $5-\frac{1}{5^9}$

14. 14.

    1. (a) $\frac{v_{n+1}}{v_n}=\frac{1}{2}$

    2. (b) $\left(v_n\right)$ is decreasing

    3. (c) $0<v_n\le 5$

    4. (d) $\frac{635}{256}$

15. 15. $p=52$

16. 18.

    1. (a) $-\infty$

    2. (b) $+\infty$

    3. (c) $+\infty$

    4. (d) $\frac{3}{2}$

    5. (e) 0

    6. (f) 0

    7. (g) $\frac{1}{2}$

    8. (h) 0

    9. (i) 3

    10. (j) 0

    11. (k) 0

    12. (l) $+\infty$

17. 19. 0

18. 20.

    1. (b) $\frac{390625}{4}$

19. 22. They are all decreasing sequences.

20. 23.

    1. (a) Show that $w_{n+1}-w_n<0$

    2. (b) Show that $0<w_n<\frac{1}{n}$ and apply the Squeeze Theorem

21. 25. Converges to 3

22. 26. Use the fact that ${\sum }_{k=1}^n{k}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

23. 27.

    1. (a) $e^{-1}$

    2. (b) 0

    3. (c) 1

    4. (d) $+\infty$

    5. (e) $+\infty$

    6. (f) 1

    7. (g) 0

    8. (h) $+\infty$

    9. (i) 9

24. 31. $lim{w}_n=2$

25. 36. 17

Series

1. 1.

   1. (a) Convergent

   2. (b) Divergent

   3. (c) Convergent

   4. (d) Divergent

   5. (e) Convergent

2. 2.

   1. (a) Convergent; $S=\frac{1}{2}$

   2. (b) Convergent; $S=1$

   3. (c) Convergent; $S=\frac{3}{2}$

   4. (d) Convergent; $S=\frac{11}{18}$

   5. (e) Convergent; $S=\frac{5}{16}$

   6. (f) Convergent; $S=3$

   7. (g) Convergent; $S=\frac{9}{8}$

   8. (h) Convergent; $S=\frac{1}{12}$

   9. (i) Divergent; $lim{a}_n=1$

3. 4. $u_n=\frac{1}{n\left(n+1\right)}$

4. 5.

   1. (a) $a_n=2n$. The series diverges since $lim{a}_n=+\infty$

   2. (b) $a_n=2{\left(-\frac{1}{3}\right)}^{n-1}$. The series converges and its sum is $S=\frac{3}{2}$

   3. (c) $a_n=\frac{1}{\left(2n-1\right)\left(2n+1\right)}$. The series converges and its sum is $S=\frac{1}{2}$

   4. (d) $a_n={\left(\frac{1}{3}\right)}^n$. The series converges and its sum is $S=\frac{3}{2}$

5. 7.

   1. (a) Convergent

   2. (b) Convergent

   3. (c) Divergent

   4. (d) Convergent

   5. (e) Convergent

   6. (f) Divergent

6. 9.

   1. (a) Convergent

   2. (b) Divergent

   3. (c) Convergent

   4. (d) Divergent

   5. (e) Divergent

   6. (f) Convergent

   7. (g) Convergent

   8. (h) Convergent

7. 10.

   1. (a) Convergent

   2. (b) Convergent

   3. (c) Convergent

   4. (d) Divergent

   5. (e) Divergent

   6. (f) Convergent

   7. (g) Convergent

   8. (h) Convergent

   9. (i) Convergent

8. 11.

   1. (a) Divergent

   2. (b) Divergent

   3. (c) Convergent

   4. (d) Convergent

   5. (e) Divergent

   6. (f) Convergent

   7. (g) Convergent

9. 12.

   1. (a) Divergent

   2. (b) Divergent

   3. (c) Convergent

   4. (d) Divergent

   5. (e) Divergent

   6. (f) Convergent

   7. (g) Divergent

   8. (h) Convergent

   9. (i) Convergent

   10. 13. (a) Divergent

   11. (b) Divergent

   12. (c) Divergent

   13. (d) Divergent

   14. (e) Divergent

10. 15.

    1. (a) Absolutely convergent

    2. (b) Conditionally convergent

    3. (c) Absolutely convergent

    4. (d) Absolutely convergent

    5. (e) Absolutely convergent

11. 16.

    1. (a) Divergent

    2. (b) Divergent

    3. (c) Absolutely convergent

    4. (d) Absolutely convergent

    5. (e) Absolutely convergent

    6. (f) Absolutely convergent

    7. (g) Divergent

    8. (h) Absolutely convergent

12. 17.

    1. (a) Pointwise convergent

    2. (b) Uniformly convergent

    3. (c) Uniformly convergent

    4. (d) Uniformly convergent

    5. (e) Uniformly convergent

13. 26.

    1. (a) $R=+\infty$ and $I=]-\infty ,+\infty [$

    2. (b) $R=\frac{1}{5}$ and $I=\left]-\frac{1}{5},\frac{1}{5}\right[$

    3. (c) $R=1$ and $I=[-1,1[$

    4. (d) $R=2$ and $I=[-2,2[$

    5. (e) $R=1$, and $I=]-1,1[$

    6. (f) $R=1$ and $I=\left[-1,1\right]$

14. 30.

    1. (a) $I=]1,5[$

    2. (b) $I=]-\infty ,+\infty [$

    3. (c) $I=[-1,1[$

    4. (d) $I=]0,4[$

    5. (e) $I=]-1,1[$

15. 31.

    1. (a) $D=]-1,1[$

    2. (b) $D=]-1,1]$

    3. (c) $D=\mathbb{R}$

    4. (d) $D=]-3,3[$

    5. (e) $D=]-2,2]$

16. 32.

    1. (a) ${\sum }_{n=0}^{+\infty }\frac{{\left(-1\right)}^n{x}^{2n}}{n!}$

    2. (b) 5

17. 33.

    1. (a) ${\sum }_{n=1}^{+\infty }\frac{{\left(-1\right)}^{n+1}{2}^{2n-1}{\left(x-4\right)}^{2n-1}}{\left(2n-1\right)!}$

    2. (d) $\frac{14}{15}$

Acknowledgments

Luís Vieira was partially supported by CMUP (UID/MAT/00144/2013), which is funded by FCT (Portugal) with national (MEC) and European structural funds through the programs FEDER under the partnership agreement PT2020.