Calculating $\sum_{n=1}^∞ \frac{1}{(2 n-1)^2+(2 n+1)^2}$ using fourier series of $\sin x$ I have to calculate $\frac{1}{1^2+3^2}+\frac{1}{3^2+5^2}+\frac{1}{5^2+7^2}+...$ using half range Fourier series $f(x)=\sin x$ which is:

$f(x)=\frac{2}{\pi}-\frac{2}{\pi}\sum_{n=2}^\infty{\frac{1+\cos
 n\pi}{n^2-1}\cos nx}$

I have no idea how to proceed. I'll appreciate if someone guide me.
 A: A different approach. Since $\frac{1}{(2n-1)^2+(2n+1)^2}=\frac{1}{8n^2+2}$ we have:
$$\begin{eqnarray*} \sum_{n\geq 1}\frac{1}{(2n-1)^2+(2n+1)^2}&=&\frac{1}{8}\sum_{n\geq 1}\frac{1}{n^2+\frac{1}{4}}\\&=&\frac{1}{8}\int_{0}^{+\infty}\sum_{n\geq 1}\frac{\sin(nx)}{n}e^{-x/2}\,dx\\&=&\frac{1}{8(1-e^{-\pi})}\int_{0}^{2\pi}\frac{\pi-x}{2}e^{-x/2}\,dx\\&=&\color{red}{\frac{1}{8}\left(-2+\pi\coth\frac{\pi}{2}\right)}.\end{eqnarray*} $$
We exploited $\int_{0}^{+\infty}\sin(ax)e^{-bx}\,dx = \frac{a}{a^2+b^2}$ and the fact that $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ is the Fourier series of a sawtooth wave.
A: Since $f(0)=\sin 0 =0$ and $\cos nx=1$ at $x=0$:
$$0=\frac{2}{\pi}-\frac{2}{\pi}\sum_{n=2}^\infty\frac{1+\cos n\pi}{n^2-1}$$
also note that $\cos n\pi=(-1)^n$. Hence
$$1=\sum_{n=2}^\infty\frac{1+\cos n\pi}{n^2-1}=\sum_{n=2}^\infty\frac{1+(-1)^n}{n^2-1}$$
All you need to know is
$$1+(-1)^n=\begin{cases}2&n=2k\\0& n=2k-1\end{cases}\,,\;k=1,2,...$$
Then
$$1=\sum_{k=1}^\infty\frac{2}{(2k)^2-1}=\sum_{k=1}^\infty\frac{4}{8k^2-2}= 4\sum_{k=1}^\infty\frac{1}{(2k-1)^2+(2k+1)^2-4}$$
