Determine convergence when AST does not work Consider this series --
$$\frac{1}{2^3}-\frac{1}{3^2}+\frac{1}{4^3}-\frac{1}{5^2}+\frac{1}{6^3}-\frac{1}{7^2}+\frac{1}{8^3}-\frac{1}{9^2}+\dots$$
Apparently, alternating series test (AST) cannot be applied. What other methods should I consider in order to determine if the series converges absolutely, converges conditionally, or diverges? Thanks.

Generically, if a series converges absolutely, then it converges (i.e. without absolute values around each term). So you might consider the absolute version,
$$ \frac{1}{2^3} + \frac{1}{3^2} + \frac{1}{4^3} + \frac{1}{5^2} + \cdots$$
This looks very similar to some very common series. It's clear that the $n$th term is bounded above by $\frac{1}{n^2}$. So then by comparison, the absolute version of your series is less than
$$ \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6} - 1.$$
