I am trying to show that $\sum_{n=1}^{\infty} \frac{ (-1)^{n-1} }{(2n -1)^3 } = \frac{ \pi^3}{32 } $. I can use the following fact: IF $f(z)$ is meromorphic with finite number of poles none which lie on the integers, then
$$ \sum_{n= - \infty}^{\infty} (-1)^n f(n) = - \sum \left( residues \; \; of \; \; \pi \csc( \pi z ) f(z) \; \; at \; \; poles \; \; of \; \; f(z) \right) $$
Notice, back to the problem, that
$$ \sum_{n=1}^{\infty} \frac{ (-1)^{n-1} }{(2n -1)^3 } = \sum_{n=1}^{\infty} \frac{ (-1)^n}{(1-2n)^3}$$ and
$$ \sum_{n=- \infty}^{\infty} \frac{ (-1)^n}{(1-2n)^3} = 1 + \sum_{n=1}^{\infty} \frac{ (-1)^n}{(1-2n)^3} + \sum_{n= - \infty}^{-1} \frac{ (-1)^n}{(1-2n)^3} = $$
$$ = 1 + \sum_{n=1}^{\infty} \frac{ (-1)^n}{(1-2n)^3} + \sum_{n=1}^{\infty} \frac{ (-1)^n}{(1+2n)^3} $$
So, I know I can find the left hand side of above equations using the result in the first line. However, I am still suck on trying to see how to evaluate $ \sum_{n=1}^{\infty} \frac{ (-1)^n}{(1+2n)^3} $. Any suggestions? Should I add the two series on the right hand side ?