How do I evaluate $\sum\limits _{n=1}^{+\infty }\frac{(-1)^{n-1}}{(2n-1)^3}$?

Asked Jun 27 '17 at 12:01

Active Jun 27 '17 at 12:03

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I know that,

$S=\left(\displaystyle\frac{1}{1^3}+\frac{1}{5^3}+\frac{1}{9^3}\cdots\right) - \left(\displaystyle\frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}\cdots\right)$

which is why it probably holds some kind of relation to the zeta-function. I tried looking it up on the internet but couldn't find anything related to $\zeta{(2k+1)}$

I want to show that $S=\displaystyle\frac{\pi^2}{32}$

Any hints ? (not asking for a complete solution) Thanks.

edited Jun 27 '17 at 12:03

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asked Jun 27 '17 at 12:01

Mathoholic99

This is just an instance of a more general identity: https://math.stackexchange.com/questions/850442/an-interesting-identity-involving-powers-of-pi-and-values-of-etas – Jack D'Aurizio Jun 27 '17 at 13:19

How do I evaluate $\sum\limits _{n=1}^{+\infty }\frac{(-1)^{n-1}}{(2n-1)^3}$?

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