How can I prove that $\sum\limits_{i=1}^\infty \frac{1}{(2i-1)^{2}} = \frac{π^2}{8}$? I have the series $$\sum\limits_{i=1}^\infty \frac{1}{i^{2}} = \frac{π^{2}}{6}$$
How can I prove that: $$\sum\limits_{i=1}^\infty \frac{1}{(2i-1)^{2}} = \frac{π^2}{8}$$
I have been looking for it all over the web, I am certain it is out there, but I am not too familiar with the mathematical terms in English. Any links or help would be appreciated.

Write
$$
\frac{\pi^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \frac{1}{(2n)^2}  + \sum_{n=1}^\infty \frac{1}{(2n-1)^2} 
$$
(i.e., separate the sum into even and odd indices). Then observe that
$$
\sum_{n=1}^\infty \frac{1}{(2n)^2}   = \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{4}\cdot \frac{\pi^2}{6} .
$$
Can you conclude from there?
