What is the result of
$$ \sum_{i=1}^{\infty} \frac{1}{(2i-1)^2}$$
I feel that the answer should be obvious, but somehow I can't find it. The series $$ 1 + \frac{1}{9} + \frac{1}{25}\ ... $$ doesn't look familiar to any other 'known' sequence. So how would you proceed here?
Thanks!
Hints:
$$\frac{\pi^2}6=\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}\;\ldots\ldots$$
First, assuming that you know the identity, $$S=1^2+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\frac{\pi^2}{6}$$
Then, $$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}=\frac{1}{2^2}\left(1^2+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots\right)=\frac{S}{4}$$
Therefore, $$1^2+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots=S-\frac{S}{4}=\frac{3S}{4}=\frac{\pi^2}{8}$$
We could do this re-arrangement in last step since series convergent absolutely(trivial).
