
# 2.2: Infinite Series


##### Definition

Given a sequence $$\left\{a_{i}\right\}_{i=m}^{\infty},$$ let $$\left\{s_{n}\right\}_{n=m}^{\infty}$$ be the sequence defined by

$s_{n}=\sum_{i=m}^{n} a_{i}.$

We call the sequence $$\left\{s_{n}\right\}_{n=m}^{\infty}$$ an infinite series. If $$\left\{s_{n}\right\}_{n=m}^{\infty}$$ converges, we call

$s=\lim _{n \rightarrow \infty} s_{n}$

the sum of the series. For any integer $$n \geq m,$$ we call $$s_{n}$$ a partial sum of the series.

We will use the notation

$\sum_{i=1}^{\infty} a_{i}$

to denote either $$\left\{s_{n}\right\}_{n=m}^{\infty},$$ the infinite series, or $$s$$, the sum of the infinite series. Of course, if $$\left\{s_{n}\right\}_{n=m}^{\infty}$$ diverges, then we say $$\sum_{i=m}^{\infty} a_{i}$$ diverges.

##### Exercise $$\PageIndex{1}$$

Suppose $$\sum_{i=m}^{\infty} a_{i}$$ converges and $$\beta \in \mathbb{R} .$$ Show that $$\sum_{i=m}^{\infty} \beta a_{i}$$ also converges and

$\sum_{i=m}^{\infty} \beta a_{i}=\beta \sum_{i=m}^{\infty} a_{i}.$

##### Exercise $$\PageIndex{2}$$

Suppose both $$\sum_{i=m}^{\infty} a_{i}$$ and $$\sum_{i=m}^{\infty} b_{i}$$ converge. Show that $$\sum_{i=m}^{\infty}\left(a_{i}+b_{i}\right)$$ converges and

$\sum_{i=m}^{\infty}\left(a_{i}+b_{i}\right)=\sum_{i=m}^{\infty} a_{i}+\sum_{i=m}^{\infty} b_{i}.$

##### Exercise $$\PageIndex{3}$$

Given an infinite series $$\sum_{i=m}^{\infty} a_{i}$$ and an integer $$k \geq m,$$ show that $$\sum_{i=m}^{\infty} a_{i}$$ converges if and only if $$\sum_{i=k}^{\infty} a_{i}$$ converges.

##### Proposition $$\PageIndex{1}$$

Suppose $$\sum_{i=m}^{\infty} a_{i}$$ converges. Then $$\lim _{n \rightarrow \infty} a_{n}=0$$.

Proof

Let $$s_{n}=\sum_{i=m}^{n} a_{i}$$ and $$s=\lim _{n \rightarrow \infty} s_{n} .$$ Since $$a_{n}=s_{n}-s_{n-1},$$ we have $$\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left(s_{n}-s_{n-1}\right)=\lim _{n \rightarrow \infty} s_{n}-\lim _{n \rightarrow \infty} s_{n-1}=s-s=0$$. Q.E.D.

##### Exercise $$\PageIndex{4}$$

Let $$s=\sum_{i=0}^{\infty}(-1)^{n} .$$ Note that

$s=\sum_{n=0}^{\infty}(-1)^{n}=1-\sum_{n=0}^{\infty}(-1)^{n}=1-s,$

from which it follows that $$s=\frac{1}{2} .$$ Is this correct?

##### Exercise $$\PageIndex{5}$$

Show that for any real number $$x \neq 1$$,

$s_{n}=\sum_{i=0}^{n} x^{i}=\frac{1-x^{n+1}}{1-x} .$

(Hint: Note that $$\left.x^{n+1}=s_{n+1}-s_{n}=1+x s_{n}-s_{n} .\right)$$

##### Theorem $$\PageIndex{2}$$

For any real number $$x$$ with $$|x|<1$$,

$\sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}.$

Proof

If $$s_{n}=\sum_{i=0}^{n} x^{i},$$ then, by the previous exercise,

$s_{n}=\frac{1-x^{n+1}}{1-x}.$

Hence

$\sum_{n=0}^{\infty} x^{n}=\lim _{n \rightarrow \infty} s_{n}=\lim _{n \rightarrow \infty} \frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}.$

## 2.2.1 Comparison Tests

The following two propositions are together referred to as the comparison test.

##### Proposition $$\PageIndex{3}$$

Suppose $$\sum_{i=m}^{\infty} a_{i}$$ and $$\sum_{i=k}^{\infty} b_{i}$$ are infinite series for which there exists an integer $$N$$ such that $$0 \leq a_{i} \leq b_{i}$$ whenever $$i \geq N .$$ If $$\sum_{i=k}^{\infty} b_{i}$$ converges, then $$\sum_{i=m}^{\infty} a_{i}$$ converges.

Proof

By Exercise 2.2 .3 We need only show that $$\sum_{i=N}^{\infty} a_{i}$$ converges. Let $$s_{n}$$ be the nth partial sum of $$\sum_{i=N}^{\infty} a_{i}$$ and let $$t_{n}$$ be the nth partial sum of $$\sum_{i=N}^{\infty} b_{i} .$$ Now

$s_{n+1}-s_{n}=a_{n+1} \geq 0$

for every $$n \geq N,$$ so $$\left\{s_{n}\right\}_{n=N}^{\infty}$$ is a nondecreasing sequence. Moreover,

$s_{n} \leq t_{n} \leq \sum_{i=N}^{\infty} b_{i}<+\infty$

for every $$n \geq N .$$ Thus $$\left\{s_{n}\right\}_{n=N}^{\infty}$$ is a nondecreasing, bounded sequence, and so converges.

##### Proposition $$\PageIndex{4}$$

Suppose $$\sum_{i=m}^{\infty} a_{i}$$ and $$\sum_{i=k}^{\infty} b_{i}$$ are infinite series for which there exists an integer $$N$$ such that $$0 \leq a_{i} \leq b_{i}$$ whenever $$i \geq N .$$ If $$\sum_{i=k}^{\infty} a_{i}$$ diverges then $$\sum_{i=m}^{\infty} b_{i}$$ diverges.

Proof

By Exercise 2.2 .3 we need only show that $$\sum_{i=N}^{\infty} b_{i}$$ diverges. Let $$s_{n}$$ be the nth partial sum of $$\sum_{i=N}^{\infty} a_{i}$$ and let $$t_{n}$$ be the nth partial sum of $$\sum_{i=N}^{\infty} b_{i} .$$ Now $$\left\{s_{n}\right\}_{n=N}^{\infty}, N$$ is a nondecreasing sequence which diverges, and so we must have $$\lim _{n \rightarrow \infty} s_{n}=+\infty .$$ Thus given any real number $$M$$ there exists an integer $$L$$ such that

$M<s_{n} \leq t_{n}$

whenever $$n>L .$$ Hence $$\lim _{n \rightarrow \infty} t_{n}=+\infty$$ and $$\sum_{i=m}^{\infty} b_{i}$$ diverges. (\quad\) Q.E.D.

##### Example $$\PageIndex{1}$$

Consider the infinite series

$\sum_{n=0}^{\infty} \frac{1}{n !}=1+1+\frac{1}{2}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots .$

Now for $$n=1,2,3, \dots,$$ we have

$0<\frac{1}{n !} \leq \frac{1}{2^{n-1}}.$

Since

$\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}$

converges, it follows that

$\sum_{n=0}^{\infty} \frac{1}{n !}$

converges. Moreover,

$2<\sum_{n=0}^{\infty} \frac{1}{n !}<1+\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}=1+\frac{1}{1-\frac{1}{2}}=3.$

We let

$e=\sum_{n=0}^{\infty} \frac{1}{n !}.$

##### Proposition $$\PageIndex{5}$$

$$e \notin \mathbb{Q}$$.

Proof

Suppose $$e=\frac{p}{q}$$ where $$p, q \in \mathbb{Z}^{+} .$$ Let

$a=q !\left(e-\sum_{i=0}^{q} \frac{1}{n !}\right).$

Then $$a \in \mathbb{Z}^{+}$$ since $$q! e=(q-1) ! p$$ and $$n !$$ divides $$q !$$ when $$n \leq q .$$ At the same time

\begin{aligned} a &=q ! \left(\sum_{n=0}^{\infty} \frac{1}{n !}-\sum_{i=0}^{q} \frac{1}{n !}\right) \\ &=q ! \sum_{n=q+1}^{\infty} \frac{1}{n !} \\ &=\left(\frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\frac{1}{(q+1)(q+2)(q+3)}+\cdots \right) \\ &= \frac{1}{q+1} \left(1+\frac{1}{q+2}+\frac{1}{(q+2)(q+3)}+\cdots\right) \\ &=\frac{1}{q+1} \left(1+\frac{1}{q+2}+\frac{1}{(q+2)(q+3)}+\cdots\right) \\ &<\frac{1}{q+1}\left(1+\frac{1}{q+1}+\frac{1}{(q+1)^{2}}+\cdots\right) \\ &=\frac{1}{q+1} \sum_{n=0}^{\infty} \frac{1}{(q+1)^{n}} \\ &=\frac{1}{q+1}\left(\frac{1}{1-\frac{1}{q+1}}\right) \\ &=\frac{1}{q}\end{aligned}

Since this is impossible, we conclude that no such integers $$p$$ and $$q$$ exist. $$\quad$$ Q.E.D.

##### Definition

We call a real number which is not a rational number an irrational number.

##### Example $$\PageIndex{2}$$

We have seen that $$\sqrt{2}$$ and $$e$$ are irrational numbers.

##### Proposition $$\PageIndex{6}$$

Suppose $$\sum_{i=m}^{\infty} a_{i}$$ and $$\sum_{i=k}^{\infty} b_{i}$$ are infinite series for which there exists an integer $$N$$ and a real number $$M>0$$ such that $$0 \leq a_{i} \leq M b_{i}$$ whenever $$i \geq N .$$ If $$\sum_{i=k}^{\infty} b_{i}$$ converges, then $$\sum_{i=m}^{\infty} a_{i}$$ converges.

Proof

Since $$\sum_{i=k}^{\infty} M b_{i}$$ converges whenever $$\sum_{i=k}^{\infty} b_{i}$$ does, the result follows from the comparison test. $$\quad$$ Q.E.D.

##### Exercise $$\PageIndex{6}$$

Suppose $$\sum_{i=m}^{\infty} a_{i}$$ diverges. Show that $$\sum_{i=m}^{\infty} \beta a_{i}$$ diverges for any real number $$\beta \neq 0$$.

##### Proposition $$\PageIndex{7}$$

Suppose $$\sum_{i=m}^{\infty} a_{i}$$ and $$\sum_{i=k}^{\infty} b_{i}$$ are infinite series for which there exists an integer $$N$$ and a real number $$M>0$$ such that $$0 \leq a_{i} \leq M b_{i}$$ whenever $$i \geq N .$$ If $$\sum_{i=m}^{\infty} a_{i}$$ diverges, then $$\sum_{i=k}^{\infty} b_{i}$$ diverges.

Proof

By the comparison test, $$\sum_{i=m}^{\infty} M b_{i}$$ diverges. Hence, by the previous exercise, $$\sum_{i=m}^{\infty} b_{i}$$ also diverges.

We call the results of the next two exercises, which are direct consequences of the last two propositions, the limit comparison test.

##### Exercise $$\PageIndex{7}$$

Suppose $$\sum_{i=m}^{\infty} a_{i}$$ and $$\sum_{i=m}^{\infty} b_{i}$$ are infinite series for which $$a_{i} \geq 0$$ and $$b_{i}>0$$ for all $$i \geq m .$$ Show that if $$\sum_{i=m}^{\infty} b_{i}$$ converges and

$\lim _{i \rightarrow \infty} \frac{a_{i}}{b_{i}}<+\infty ,$

then $$\sum_{i=m}^{\infty} a_{i}$$ converges.

##### Exercise $$\PageIndex{8}$$

Suppose $$\sum_{i=m}^{\infty} a_{i}$$ and $$\sum_{i=m}^{\infty} b_{i}$$ are infinite series for which $$a_{i} \geq 0$$ and $$b_{i}>0$$ for all $$i \geq m .$$ Show that if $$\sum_{i=m}^{\infty} a_{i}$$ diverges and

$\lim _{i \rightarrow \infty} \frac{a_{i}}{b_{i}}<+\infty ,$

then $$\sum_{i=m}^{\infty} b_{i}$$ diverges.

##### Exercise $$\PageIndex{9}$$

Show that

$\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}$

converges.

##### Exercise $$\PageIndex{10}$$

Show that

$\sum_{n=0}^{\infty} \frac{x^{n}}{n !}$

converges for any real number $$x \geq 0$$.

##### Exercise $$\PageIndex{11}$$

Let $$S$$ be the set of all finite sums of numbers in the set $$\left\{a_{1}, a_{2}, a_{3}, \dots\right\},$$ where $$a_{i} \geq 0$$ for $$i=1,2,3, \dots$$ That is

$S=\left\{\sum_{i \in J} a_{i}: J \subset\{1,2,3, \ldots, n\} \text { for some } n \in \mathbb{Z}^{+}\right\}.$

Show that $$\sum_{i=1}^{\infty} a_{i}$$ converges if and only if $$\sup S<\infty,$$ in which case

$\sum_{i=1}^{\infty} a_{i}=\sup S.$

One consequence of the preceding exercise is that the sum of a sequence of nonnegative numbers depends only on the numbers begin added, and not on the order in which they are added. That is, if $$\varphi: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$$ is one-to-one and onto, $$\sum_{i=1}^{\infty} a_{i}$$ converges if and only if $$\sum_{i=1}^{\infty} a_{\varphi(i)}$$ converges, and, in that case,

$\sum_{i=1}^{\infty} a_{i}=\sum_{i=1}^{\infty} a_{\varphi(i)}.$