Series

What is

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

What is

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

What is

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

Consider the following sequence of partial sums:

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

Can we compute

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

These partial sums look as follows:

$\displaystyle 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 $ \displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n} = 1$, if it makes any sense. But does it?

In a moment we will define

$\displaystyle \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

$\displaystyle \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

$\displaystyle 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

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

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

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

as claimed earlier.

William Stein 2006-03-15