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2.1 Series versus sequence

Given a sequence $(a_{n})_{n\in\mathbb{N}}$ in $\mathbb{R}$ , we can write

a_{1}+a_{2}+\ldots+a_{n}=\sum_{i=1}^{n}a_{i},

for every fixed $n\in\mathbb{N}$ : we take the sum of the first $n$ terms of the sequence by induction. But can we also study expressions of the form

a_{1}+a_{2}+a_{3}+\ldots,

where the implication is that the addition goes on forever? What does this mean? Summing up all terms of the sequence? Nobody (and no computer) can actually perform infinitely many additions. In the same way as $a_{\infty}$ does not exist as a term of the sequence $(a_{n})_{n\in\mathbb{N}}$ , we cannot sum up infinitely many terms. Consider an example:

\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots.

Here the “running total” is successively $\frac{1}{2}$ , $\frac{3}{4}$ , $\frac{7}{8}$ , $\frac{15}{16}$ , …, and these numbers tend to $1$ according to MATH113. What mathematical concept does this remind you of?

Sums have always finitely many summands, so we need to properly define what we mean by these “infinite sums” in general. How to do this? And how to compute it afterwards? Given $a_{1}$ , $a_{2}$ , $a_{3}$ , …in $\mathbb{R}$ , let

A_{n}:=a_{1}+a_{2}+\cdots+a_{n}=\sum_{i=1}^{n}a_{i},

be the so-called $n$ -th partial sum. If the sequence $(A_{n})_{n\in\mathbb{N}}$ tends to a limit as $n\to\infty$ , we could say

a_{1}+a_{2}+a_{3}+\cdots:=\lim_{n\rightarrow\infty}A_{n}.

We formalise this in the following definition:

Definition 2.1.1.

Given a sequence $(a_{n})_{n\in\mathbb{N}}$ in $\mathbb{R}$ , the corresponding series is the formal expression $\sum_{i=1}^{\infty}a_{n}=a_{1}+a_{2}+a_{3}+\ldots$ .

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The series is said to be convergent if the sequence of partial sums $\big{(}\sum_{i=1}^{n}a_{i}\big{)}_{n\in\mathbb{N}}$ converges, and in this case we call this limit the sum of the series and write

$\sum_{i=1}^{\infty}a_{i}=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}a_{i}\in% \mathbb{R},$

so we can identify the series with a real number;
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otherwise, the series $\sum_{i=1}^{\infty}a_{i}$ is divergent and remains a formal expression with no associated real number.

Comment: Be careful: like a sequence, a series may converge or diverge, but so far we have no connection between the convergence/divergence of a sequence and the convergence/divergence of the corresponding series!

We can relate these terms as follows:

original sequence $(a_{n})_{n\in\mathbb{N}}$

$\updownarrow$

sequence of partial sums $(A_{n})_{n\in\mathbb{N}}=(\sum_{i=1}^{n}a_{i})_{n\in\mathbb{N}}$

$\updownarrow$

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

You can move forward and backward between them.

Example 2.1.2.

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Given a sequence like $\displaystyle(\frac{1}{2^{n}})_{n\in\mathbb{N}}$ , the corresponding sequence of partial sums is determined by

$A_{n}=\frac{1}{2}+\ldots+\frac{1}{2^{n}}=\frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}% {2}}-1,\quad n\in\mathbb{N}.$

The last equality is proved by induction, MATH101. We know from MATH113 that $A_{n}\rightarrow\frac{1}{1-\frac{1}{2}}-1=1$ as $n\rightarrow\infty$ , so the corresponding series $\sum_{i=1}^{\infty}\frac{1}{2^{i}}$ converges and $\sum_{i=1}^{\infty}\frac{1}{2^{i}}=1$ .
•

Given a series like $\displaystyle\sum_{i=1}^{\infty}\frac{1}{i}$ , the corresponding original sequence is $(\frac{1}{n})_{n\in\mathbb{N}}$ , and the sequence of partial sums $(1+\frac{1}{2}+\ldots+\frac{1}{n})_{n\in\mathbb{N}}$ .
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Given a sequence of partial sums $\displaystyle(A_{n})_{n\in\mathbb{N}}=\Big{(}\frac{1}{n}\Big{)}_{n\in\mathbb{N}}$ , the corresponding original sequence must have been $(a_{n})_{n\in\mathbb{N}}$ , where $a_{1}=1$ and $a_{n}=\frac{1}{n}-\frac{1}{n-1}$ , for all $n\geq 2$ . This is so because $A_{n+1}=A_{n}+a_{n+1}$ , so $a_{n+1}=A_{n+1}-A_{n}=\frac{1}{n+1}-\frac{1}{n}$ . Since the sequence of partial sums

$(A_{n})_{n\in\mathbb{N}}=\Big{(}\sum_{i=1}^{n}a_{i}\Big{)}_{n\in\mathbb{N}}=% \Big{(}\frac{1}{n}\Big{)}_{n\in\mathbb{N}}$

converges to $0$ , the corresponding series $\sum_{i=1}^{\infty}a_{i}$ converges and $\sum_{i=1}^{\infty}a_{i}=0$ . Notice that we have not said anything about the series $\sum_{n=1}^{\infty}\frac{1}{n}$ and in fact it will be shown in Proposition 2.2.4 that the latter one diverges.
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Some more examples:

Now some elementary facts that follow at once from the corresponding statements about limits:

Proposition 2.1.3.

Suppose we have two convergent series $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ . Then:

$(i)$

$\displaystyle\sum_{n=1}^{\infty}ca_{n}=c\sum_{n=1}^{\infty}a_{n}$ for every constant $c\in\mathbb{R}$ ;
$(ii)$

$\displaystyle\sum_{n=1}^{\infty}(a_{n}+b_{n})=\sum_{n=1}^{\infty}a_{n}+\sum_{n% =1}^{\infty}b_{n}$ ;
$(iii)$

if $a_{n}\leqslant b_{n}$ for all $n\in\mathbb{N}$ then $\sum_{n=1}^{\infty}a_{n}\leqslant\sum_{n=1}^{\infty}b_{n}$ .

Proof.

Exercise X. ∎

Proposition 2.1.4.

If a series is convergent/divergent then so is any other series formed from it by altering only finitely many summands.

Proof.

Exercise A2.1. ∎

Proposition 2.1.5 (Basic non-convergence test).

Let $(a_{n})_{n\in\mathbb{N}}$ be a sequence. Then

if $a_{n}\not\rightarrow 0$ as $n\rightarrow\infty$ then the series $\sum_{n=1}^{\infty}a_{n}$ is divergent.

Comment: This is useful in order to tell with certainty if a series diverges – but it does not always detect divergence.

Proof.

We prove the contrapositive: if $\sum_{n=1}^{\infty}a_{n}$ converges then $a_{n}\rightarrow 0$ as $n\rightarrow\infty$ . In fact, notice that, if the series converges then by definition the sequence of partial sums $(A_{n})_{n\in\mathbb{N}}$ converges, say to some number $A$ , so

a_{n}=A_{n}-A_{n-1}\rightarrow A-A=0,\quad n\rightarrow\infty,

proving our claim. ∎

Example 2.1.6.

Let $a_{n}=\displaystyle\frac{n-1}{2n+1}$ . Then $\sum_{n=1}^{\infty}a_{n}$ is divergent, because

a_{n}=\frac{1-1/n}{2+1/n}\to\frac{1}{2}\neq 0\qquad\mbox{as}\quad n\to\infty.

Example 2.1.7.

Consider the series $\sum_{n=1}^{\infty}\cos n$ . The summands somehow cancel with each other so the series should converge, right? However, we see that $\cos n\not\rightarrow 0$ as $n\rightarrow\infty$ . More precisely, for every $N\in\mathbb{N}$ , you can find $n>N$ such that $\cos n>\frac{1}{2}$ . Thus the basic non-convergence test shows that $\sum_{n=1}^{\infty}\cos n$ diverges.

What about $\sum_{n=1}^{\infty}\cos(n\pi)$ instead? Well, we know that $\cos(n\pi)=(-1)^{n}$ . However, $(-1)^{n}\not\rightarrow 0$ as $n\rightarrow 0$ , so $\sum_{n=1}^{\infty}\cos(n\pi)=\sum_{n=1}^{\infty}(-1)^{n}$ diverges, too.

Exercise A3.2.

Write an essay on the relation between sequences and series. Use your own words, examples and illustrations. Be mathematically correct and precise. 200-300 words, excluding mathematical symbols or pictures.