Series

What is

$$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots?$$

What is

$$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \frac{1}{243} + \ldots?$$

What is

$$\frac{1}{1} + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \ldots?$$

Consider the following sequence of partial sums:

$$a_N = \sum_{n=1}^N \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^N}.$$

Can we compute

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

These partial sums look as follows:

$$a_1 = \frac{1}{2},\qquad a_2 = \frac{3}{4},\qquad a_{10} = \frac{1023}{1024},\qquad a_{20} = \frac{1048575}{1048576}$$

It looks very likely that $$\sum_{n=1}^{\infty} \frac{1}{2^n} = 1$$, if it makes any sense. But does it?

In a moment we will define

$$\sum_{n=1}^{\infty} \frac{1}{2^n} = \lim_{N\to\infty} \sum_{n=1}^N \frac{1}{2^n} = \lim_{N\to\infty} a_N.$$

A little later we will show that $a_{N} = \frac{2^N - 1}{2^N}$, hence indeed $\sum_{n=1}^{\infty} \frac{1}{2^n} = 1$.

Definition 6.2.1 (Sum of series)   If $\left(a_n\right)_{n=1}^{\infty}$ is a sequence, then the sum of the series is

$$\sum_{n=1}^{\infty} a_n = \lim_{N\to\infty} \sum_{n=1}^N a_n = \lim_{N\to\infty} s_N$$

provided the limit exists. Otherwise we say that $\sum_{n=1}^{\infty} a_n$ diverges.

Example 6.2.2 (Geometric series)   Consider the geometric series $\sum_{n=1}^{\infty} a r^{n-1}$ for $a\neq 0$. Then

$$s_N = \sum_{n=1}^N a r^{n-1} = \frac{a(1-r^N)}{1-r}.$$

To see this, multiply both sides by $1-r$ and notice that all the terms in the middle cancel out. For what values of $r$ does $\lim_{N\to\infty} \frac{a(1-r^N)}{1-r}$ converge? If $\vert r\vert<1$, then $\lim_{N\to\infty} r^N = 0$ and

$$\lim_{N\to\infty} \frac{a(1-r^N)}{1-r} = \frac{a}{1-r}.$$

If $\vert r\vert> 1$, then $\lim_{N\to\infty} r^N$ diverges, so $\sum_{n=1}^{\infty} a r^{n-1}$ diverges. If $r=\pm 1$, it's clear since $a\neq 0$ that the series also diverges (since the partial sums are $s_N = \pm Na$).

For example, if $a=1$ and $r=\frac{1}{2}$, we get

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

as claimed earlier.