Alternating Series Test

Alternating Series

In this section we’ll discuss alternating series, and a simple test that can be done in order to show that many alternating series converge.

An Alternating Series is a series in which every other term is positive and every other term is negative. That is, an alternating series has the form

$\sum\limits_{n=0}^\infty (-1)^n a_n$

where all $a_n$ are positive.
(Alternately, the series could start at $n=1$ , or the exponent of $-1$ could be $n+1$ or $n-1$ . These are all valid variations, allowing first term of the series to be negative.)

Example:
The series $\sum\limits_{n=0}^\infty \frac{(-1)^n}{2^n} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdot \cdot \cdot$ is an alternating series.
Looking at the first few partial sums of this series:
$S_0 = 1$
$S_1 = 1 - \frac{1}{2} = \frac{1}{2} = 0.5$
$S_2 = 1 - \frac{1}{2} + \frac{1}{4} = \frac{3}{4} = 0.75$
$S_3 = 1- \frac{1}{2} + \frac{1}{4} - \frac{1}{8} = \frac{5}{8} = 0.625$
$S_4 = 1- \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} = \frac{11}{16} = 0.6875$

It appears that this series converges. (It does, in fact, to $\frac{2}{3}$ .) But there is a quick test that can be done, in order to show that this series converges.

Alternating Series Test:
For an alternating series $\sum\limits_{n=0}^\infty (-1)^n a_n$ , if the terms $a_n$
1.) are monotone decreasing ( $a_{n+1} \leq a_n$ for all $n\geq 0$ ) and
2.) satisfy $\lim\limits_{n\rightarrow \infty} a_n = 0$ ,
then the alternating series converges.

Additionally, the partial sums $S_k$ are increasingly good approximations of the sum of the entire series, and there is an approximation for the error between a partial sum and the sum of the series:

$| \sum\limits_{n=0}^\infty (-1)^n a_n - S_k | \leq a_{k+1}$

That is: the difference between any partial sum, and the sum of the infinite series, is less than the subsequent term in the series.

Example:
The series $\sum\limits_{n=0}^\infty \frac{(-1)^n}{2^n}$ converges, because $\frac{1}{2^{n+1}} \leq \frac{1}{2^{n}}$ for all $n$ and $\lim\limits_{n \rightarrow \infty} \frac{1}{2^n} = 0$ .

Also, looking at the partial sums from before:
$S_0 = 1$
$S_1 = \frac{1}{2} = 0.5$
$S_2 = \frac{3}{4} = 0.75$
$S_3 = \frac{5}{8} = 0.625$
$S_4 = \frac{11}{16} = 0.6875$
you will notice that each partial sum is closer to $\frac{2}{3}$ than the previous term, and that the approximations are alternately higher and lower than the actual sum of $\frac{2}{3}$ . The error between $S_4 = \frac{11}{16} = 0.6875$ and $\sum\limits_{n=0}^\infty \frac{(-1)^n}{2^n} = \frac{2}{3}$ is $| \frac{2}{3} - \frac{11}{16}| = \frac{1}{48} < \frac{1}{32}= a_5$

Examples:
Determine whether the following series can be shown to converge, using the alternating series test.

$\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ converges because $\frac{1}{n+1} < \frac{1}{n}$ and $\lim\limits_{n\rightarrow \infty} \frac{1}{n} = 0$ .
(In fact, it converges to $\ln 2$ ; see the Maclaurin Series for $\ln (1+x)$ , evaluated at $x = 1$ .)

The alternating series test does not instantly apply to $\sum\limits_{n=0}^\infty \frac{(-1)^n}{n!}$ because $a_0 = 1 = a_1$ . However, this series may be re-written as $1 + \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!}$ and that series converges because $\frac{1}{n!} \leq \frac{1}{(n+1)!}$ and $\lim\limits_{n\rightarrow \infty} \frac{1}{n!} = 0$ for all $n > 0$ .
(In fact, it converges to $\frac{1}{e}$ ; see the Maclaurin Series for $e^{-x}$ , evaluated at $x = 1$ , or for $e^{x}$ , evaluated at $x = -1$ )

So the Alternating Series Test can be re-stated:
For an alternating series $\sum\limits_{n=0}^\infty (-1)^n a_n$ , if the terms $a_n$
1.) are eventually monotone decreasing ( $a_{n+1} \leq a_n$ for all $n\geq d$ , where $d$ is some positive number) and
2.) satisfy $\lim\limits_{n\rightarrow \infty} a_n = 0$ ,
then the alternating series converges.

Absolutely versus Conditionally Convergent:
A convergent alternating series $\sum\limits_{n=0}^\infty (-1)^n a_n$ is said to converge absolutely if $\sum\limits_{n=0}^\infty a_n$ also converges. Otherwise, the series is conditionally convergent.

Examples:
The series $\sum\limits_{n=0}^\infty \frac{(-1)^n}{2^n}$ is absolutely convergent, because $\sum\limits_{n=0}^\infty \frac{1}{2^n} = 2$ .

The series $\sum\limits_{n=1}^\infty \frac{(-1)^n}{n}$ is conditionally convergent, because the series $\sum\limits_{n=1}^\infty \frac{1}{n}$ diverges.

$\sum\limits_{n=0}^\infty \frac{(-1)^n}{n!}$ is an absolutely convergent series, because $\sum\limits_{n=0}^\infty \frac{1}{n!} = e$ .
(Look at the Maclaurin Series for $y = e^x$ , evaluated at $x = 1$ .)

Final Notes:
The alternating series test can readily identify many alternating series that converge. However, a series that does not satisfy the alternating series test does not necessarily diverge.

The claims: $\sum\limits_{n=0}^\infty \frac{(-1)^n}{2^n} = \frac{2}{3}$ , $\sum\limits_{n=0}^\infty \frac{1}{2^n} = 2$ , and $\sum\limits_{n=1}^\infty \frac{1}{n}$ diverges were not proven. Can you prove them?

Alternating Series Test

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