# How to get the sum of the series $\sum _{n=1}^{\infty } \frac{1}{\left(4 n^2-1\right)^2}$?

So, in order to obtain the required answer, I tried to apply some Taylor expansions, which led me to nowhere actually. After a while I tried to use the summation theorem

$\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}}$ at $f\left(z\right)$'s poles.

Residue at $z=-\frac{1}{2}$ equals to residue at $z=\frac{1}{2}$, both of them are $-\frac{\pi ^2}{16}$. What I got seems to be not the correct answer after all:

$-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}} = -\frac{\pi ^2}{8}$, so that $\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{8}$ Since the $f\left(z\right)$ is even, I get $\sum_{n=0}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{16}$

However, the initial task was to find the sum of $\sum_{n=1}^{+\infty}{f\left(n\right)}$. I supposed that $\sum_{n=1}^{+\infty}{f\left(n\right)}=\sum_{n=0}^{+\infty}{f\left(n\right)}-f\left(0\right)=\frac{\pi ^2}{16}-1$, which is incorrect, as according to Mathematica, I should get $\frac{\pi ^2}{16}-\frac{1}{2}$. I don't know where is my mistake at this point. Perhaps, the whole approach is not well executed.

Since $\frac{1}{4n^2-1}=\frac{1}{2}\left(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}\right)$ by squaring we get $$\sum_{n\geq 1}\frac{1}{(4n^2-1)^2} = \frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n-1)^2}+\frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n+1)^2}-\frac{1}{4}\sum_{n\geq 1}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$ where the first two series in the RHS depend on $\zeta(2)-\frac{1}{4}\zeta(2)=\frac{\pi^2}{8}$ and the last one is telescopic: $$\sum_{n\geq 1}\frac{1}{(4n^2-1)^2}=\color{blue}{\frac{\pi^2-8}{16}}.$$