<s>[INST] Sum of a nearly classic series Assuming we know that : $$\sum_{n=1}^{+\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}$$
How do you find the sum of a series where all terms are in this one ?
For instance, how do you prove that ?$$\sum_{n=1}^{+\infty}{\frac{1}{(2n-1)^2}} = \frac{\pi^2}{8}$$
 [/INST] Observe that:
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty}\frac{1}{(2n)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{6}$$
and 
$$\sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{24} $$
therefore
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} & = & \sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{n=1}^{\infty} \frac{1}{(2n)^2} \\
& = & \frac{\pi^2}{6} - \frac{\pi^2}{24} \\
& = & \frac{\pi^2}{8}
\end{eqnarray*}
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