Alternating series theorem fails if terms are not monotonically decreasing in magnitude

Statement

Basic statement

Consider a series of the form:

$a_{1}+a_{2}+\dots +a_{n}+\dots =\sum _{k=1}^{\infty }a_{k}$

Suppose the following three conditions hold for the series:

1. Alternating signs: All the $a_{k}$s are nonzero and the sign of $a_{k+1}$ is opposite the sign of $a_{k}$.
2. Not necessarily monotonically decreasing in magnitude: It is not necessarily true that $|a_{k}|\geq |a_{k+1}|$ for all $k$
3. Terms approach zero: $\lim _{k\to \infty }a_{k}=0$.

Then, any of the following is possible for the series:

• The series is an absolutely convergent series
• The series is a conditionally convergent series
• The series diverges to $+\infty$ or $-\infty$
• The partial sums of the series have differing values of limit superior and limit inferior.

Proof

Example of a divergent series

Here is an example of an alternating series whose terms approach zero but which diverges:

$1-{\frac {1}{3}}+{\frac {1}{2}}-{\frac {1}{6}}+{\frac {1}{3}}-{\frac {1}{9}}+{\frac {1}{4}}-{\frac {1}{12}}+\dots$

To see that the series diverges, group the terms as:

$\left(1-{\frac {1}{3}}\right)+\left({\frac {1}{2}}-{\frac {1}{6}}\right)+\dots$

The $k^{th}$ grouped term is $1/k-1/(3k)=2/(3k)$, and we know that $\sum _{k=1}^{\infty }2/(3k)={\frac {2}{3}}\sum _{k=1}^{\infty }1/k$ diverges.