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.

• Do you see any series that you could compare it to? – user296602 Jan 6 '16 at 4:18
• @user yes good idea – Forever Mozart Jan 6 '16 at 4:21

look at the series constructed of the sum of $1/(2n)^2 - 1/(2n+1)^2$. You get a sum that is converging absolutely if I am not mistaken.

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 $$\frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6} - 1.$$