<s>[INST] Proving that the sum $\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$ Proving that the sum

$$\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$$
  When $m$=integer even number

I know that the $\frac{\pi^2}{8}$ comes from $\sum_{n=0}^{\infty }\frac{1}{(2n+1)^2}$ and I know how to prove it but I don't know how to prove the above sum  especially with the power $m$
 [/INST] $\sin(n\frac{\pi}{2}) \in \{-1, 0, 1\}$
For even $m$,
$\sin^m(n\frac{\pi}{2}) = \begin{cases}
0 : n \text{ is even} \\
1 : n \text{ is odd}
\end{cases}$
So, we can rewrite this summation as:
$$
\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2} \\
= \sum_{n=1}^{\infty }\frac{1}{n^2} - \sum_{n=1}^{\infty }\frac{1}{(2n)^2} \\
= \sum_{n=1}^{\infty }\frac{1}{n^2} - \frac{1}{4}\sum_{n=1}^{\infty }\frac{1}{n^2} \\
= \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} \\
= \frac{\pi^2}{8}
$$
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