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 $m$
Let $m=2n$, where $n\in\mathbb{N}^{+}$. IOW, $m$ is a positive even number (that statement is false for $m=0$).
$$\begin{align} s_{2n} &=\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(\frac{k\pi}{2}\right)}}{k^2}\\ &=\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(\frac{(2k-1)\pi}{2}\right)}}{(2k-1)^2}+\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(\frac{(2k)\pi}{2}\right)}}{(2k)^2};~~\text{(split into even/odd terms)}\\ &=\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(\frac{(2k-1)\pi}{2}\right)}}{(2k-1)^2}\color{red}{+\sum_{k=1}^{\infty}\frac{\sin^{2n}{\left(k\pi\right)}}{4k^2}};~~\text{(even terms vanish)}\\ &=\sum_{k=0}^{\infty}\frac{\sin^{2n}{\left(\frac{(2k+1)\pi}{2}\right)}}{(2k+1)^2}\\ &=\sum_{k=0}^{\infty}\frac{\left[\sin{\left(k\pi+\frac{\pi}{2}\right)}\right]^{2n}}{(2k+1)^2}\\ &=\sum_{k=0}^{\infty}\frac{\left[\cos{\left(k\pi\right)}\right]^{2n}}{(2k+1)^2};~~~[\sin{(z+\pi/2)}=\cos{z}]\\ &=\sum_{k=0}^{\infty}\frac{\left[(-1)^{k}\right]^{2n}}{(2k+1)^2}\\ &=\sum_{k=0}^{\infty}\frac{1^{n}}{(2k+1)^2}\\ &=\frac{\pi^2}{8}.~~\blacksquare\\ \end{align}$$
$\sin(n\frac{\pi}{2}) \in \{-1, 0, 1\}$
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:
$$ \sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2} \\ = \sum_{n=1}^{\infty }\frac{1}{n^2} - \sum_{n=1}^{\infty }\frac{1}{(2n)^2} \\ = \sum_{n=1}^{\infty }\frac{1}{n^2} - \frac{1}{4}\sum_{n=1}^{\infty }\frac{1}{n^2} \\ = \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} \\ = \frac{\pi^2}{8} $$
observe that $\sin\frac{n\pi}{2}$ for $n=0,1,2,3,\dots$ is always $0$ or $1$ or $-1$
Your sum convereges absolutely, so you can partition it into 2 parts: part where the numerator is 1 and part where the numerator is $-1$.
Hint: Determine the expression $\sin^m{(n\pi/2)}$. From this point of view you can do some machinary work for solving the series.
After some calculations you get the series $$\sum _{k=1}^{\infty } \frac{1-(-1)^k}{2k^2}$$
Finally you have to compute the series. Maybe some Fourier series are useful or you are able to split the series in smart partial series.
Hint : $\sin^m(n\frac{\pi}{2})= (-1)^{(n+1).m}$ or $0$ for $n=2k$
