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 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.
