5.1 Series

Outside the world of mathematics the words ‘sequence’ and ‘series’ can often mean the same thing, however, within the context of real analysis this is not the case. We define a series to be an infinite summation, which we can think of as a sum of all of the terms in a sequence. For example, we can take any sequence $(x_{n})_{n}$ and form the series

\sum_{n=1}^{\infty}x_{n}=x_{1}+x_{2}+x_{3}+\dots.

(5.1)

But when is it valid to write that a series has some finite value? If we add up infinitely many things there will be many cases where the sum is infinite, for example a series constructed from any constant sequence would just be that constant added up infinitely many times, which obviously has no finite value. To solve this problem, we introduce a new tool: the sequence of partial sums.

Definition 5.1.

Let $S$ be a series $\sum_{n=1}^{\infty}x_{n}$ generated by some sequence $(x_{n})_{n}$ . Then we define the $n$ ^th partial sum of $S$ as

S_{n}=\sum_{k=1}^{n}=x_{1}+x_{2}+\dots+x_{n}.

(5.2)

5.1.1 Convergence of Series

It is clear that the $n$ ^th partial sums always have a finite value (if the elements of the sequence are well-defined) since they are only finite sums. This means we can consider the partial sums as a sequence in their own right $(S_{n})_{n}$ , and we can say that if the sequence of partial sums converges to some value, then the infinite series takes on that exact value.

Definition 5.2.

Let $S$ be a series $\sum_{n=1}^{\infty}x_{n}$ . Then we say $S$ converges if $(S_{n})_{n}$ converges, in which case

S=\sum_{n=1}^{\infty}x_{n}=\lim_{n\to\infty}S_{n}.

(5.3)

If $(S_{n})_{n}$ diverges, then $\sum_{n=1}^{\infty}$ diverges and $S$ is either not defined, or we can say that $S=\infty$ or $S=-\infty$ , depending on the nature of the divergence of $(S_{n})_{n}$ .

Example 5.1.

Consider the sequence given by $x_{n}=\frac{1}{n(n+1)}$ . The sequence of partial sums of the series generated by this sequence is given by

$\displaystyle S_{n}$	$\displaystyle=\sum_{k=1}^{n}\frac{1}{k(k+1)}$	(5.4)
	$\displaystyle=\frac{1}{1\cdot}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+% \frac{1}{n(n+1)}$	(5.5)
	$\displaystyle=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{% 3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{n}-\frac{1% }{n+1}\right)$	(5.6)
	$\displaystyle=1-\frac{1}{n+1}.$	(5.7)

We can see that $(S_{n})_{n}$ converges to 1, and hence $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is a convergent series with the value 1.

Example 5.2.

Consider the seqeuence given by $x_{n}=(-1)^{n}$ . Then

S_{1}=-1,\ S_{2}=-1+(-1)^{2}=0,\ S_{3}=-1+(-1)^{2}+(-1)^{3}=-1\dots

(5.8)

Thus the $n$ th partial sum is given by

S_{n}=\begin{cases}-1&\text{if $n$ is odd}\\ 0&\text{if $n$ is even}\end{cases}

(5.9)

This sequence $(S_{n})_{n}$ is divergent, and so the series $\sum_{n=1}^{\infty}(-1)^{n}$ is not defined.

5.1.2 Convergence of the Generating Sequence

In the first example we can see that the sequence which generated the convergent series converged to 0 (not the partial sums!). This turns out to be a general fact which we will now show.

Theorem 5.1

Let $S=\sum_{n=1}^{\infty}x_{n}$ be a convergent series. Then $x_{n}\to 0$ as $n\to\infty$ .

Proof.

Since $S$ is convergent, the sequence $(S_{n})_{n}$ converges. Note that $x_{n}=S_{n}-S_{n-1}$ for all $n\geq 2$ , and since both $S_{n}\to S$ and $S_{n-1}\to S$ as $n\to\infty$ , we have

x_{n}\to S-S=0\ \text{as}\ n\to\infty.

(5.10)

∎

It makes sense that this should be the case. If the generating sequence did not tend to 0, then we would be adding together infinitely many nonzero terms, and it would be very strange if this somehow converges! It would be useful if the converse of this theorem was also true ( $x_{n}\to 0\implies\sum_{n=1}^{\infty}$ converges), but this turns out not to be the case and leads to one of the most (in)famous counterexamples in mathematics.

Example 5.3 (The Harmonic Series).

The series $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges to $+\infty$ .
Fist note that $(S_{n})_{n}$ is an increasing sequence since $S_{n}-S_{n-1}=\left(\sum_{k=1}^{n}\frac{1}{k}\right)-\left(\sum_{k=1}^{n-1}% \frac{1}{k}\right)=\frac{1}{n}\geq 0$ . Now consider the subsequence $(S_{2^{m}})_{m}$ . The general term of this sequence is given by

	$\displaystyle S_{2^{m}}=$	$\displaystyle 1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1% }{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\dots+\frac{% 1}{16}\right)$		(5.11)
		$\displaystyle+\dots+\left(\frac{1}{2^{m-1}+1}+\dots+\frac{1}{2^{m}}\right),$		(5.12)

where the brackets indicate how the terms are added on in each partial sum. The $i$ th bracketed expression in each term is given by

t_{i}=\frac{1}{2^{i-1}+1}+\frac{1}{2^{i-1}+2}+\dots+\frac{1}{2^{i}}.

(5.13)

There are $2^{i}-2^{i-1}$ values in this sum, and they are all greater than or equal to $\frac{1}{2^{i}}$ , so $t_{i}\geq 2^{i-1}\frac{1}{2^{i}}=\frac{1}{2}$ . Hence

S_{2^{m}}=1+t_{1}+t_{2}+\dots+t_{m}\geq 1+\frac{m}{2}.

(5.14)

Now, let $K\in\mathbb{N}$ , then for $n\geq 2^{2K}$ we have

S_{n}\geq S_{2^{2K}}\geq 1+\frac{2K}{2}=1+K>K,

(5.15)

and so since the sequence of partials sums diverges to $\infty$ , the series must also diverge.

5.1.3 Geometric Series

Definition 5.3.

A geometric series is an infinite series of terms which have a constant ratio between them. They can be expressed as

\sum_{n=1}^{\infty}ar^{n-1}

(5.16)

for some $a\in\mathbb{R}$ and $r\in\mathbb{R}\setminus\{0\}$ .

Theorem 5.2

The geometric series $\sum_{n=1}^{\infty}ar^{n-1}$ converges if and only if $\lvert r\rvert<1$ , and in this case $\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r}$ .

Proof.

First note the general term of the sequence of partial sums:

S_{n}=\sum_{k=1}^{n}ar^{k-1}=a+ar+ar^{2}+\dots+ar^{k-1}=\frac{a(1-r^{n})}{1-r}.

(

r\neq 1

)

If $\lvert r\rvert<1$ , then by theorem 4.1 $r^{n}\to 0$ as $n\to\infty$ . Hence $S_{n}\to\frac{a}{1-r}$ . If $r=1$ , then $S_{n}=an$ which tends to $\pm\infty$ depending on the sign of $a$ . If $r=-1$ , then $S_{n}$ is 0 if $n$ is even and $a$ if $n$ is odd, hence $S_{n}$ diverges. It can be shown that if $\lvert r\rvert>1$ , $S_{n}$ diverges, hence all the cases are covered and the only way the series can converge is if $\lvert r\rvert<1$ . ∎