Calculate $\sum_{n=0}^\infty \frac1{(4n^2 - 1)^2}$ How do I find the value of the following infinite series?

$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} $$

My attempt at a solution:
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} = \sum_{n=0}^\infty \frac{1}{((2n-1)(2n+1))^2} = \sum_{n=0}^\infty \left(\frac{1}{2}\left(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}\right)\right)^2  = \frac{1}{4}\sum_{n=0}^\infty \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)^2 $$ 
I then tried to compute the partial sums of this series, but with no luck. Does anyone else know how to do it?
 A: Note that$$(\forall n\in\mathbb{N}):\frac1{(4n^2-1)^2}=\frac1{4(2n+1)}-\frac1{4 (2n-1)}+\frac1{4(2n+1)^2}+\frac1{4(2n-1)^2}.$$It is clear that the series$$\sum_{n=0}^\infty\frac1{4(2n+1)}-\frac1{4(2n-1)}$$is a telescoping series, whose sum is $\frac14$. On the other hand\begin{align}\sum_{n=0}^\infty\frac{1}{4 (2 n+1)^2}+\frac{1}{4(2 n-1)^2}&=\frac14+2\sum_{n=0}^\infty\frac{1}{4 (2 n+1)^2}\\&=\frac14+\frac12\sum_{n=0}^\infty\frac1{(2n+1)^2}.\end{align}But, since$$\sum_{n=0}^\infty\frac1{(2n+1)^2}=\frac{\pi^2}8,$$we have that the sum of your series is$$\frac14+\frac14+\frac{\pi^2}{16}=\frac12+\frac{\pi^2}{16}.$$
A: Since 
$$ \left|\sin x\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1} $$
by Parseval's theorem we have
$$ \pi=\int_{-\pi}^{\pi}\left|\sin x\right|^2\,dx =2\pi\cdot\frac{4}{\pi^2}+\frac{16}{\pi}\sum_{n\geq 1}\frac{1}{(4n^2-1)^2}$$
hence by rearranging:
$$ \sum_{n\geq 1}\frac{1}{(4n^2-1)^2} = \color{red}{\frac{\pi^2}{16}-\frac{1}{2}}.$$
A: Let's continue with the approach you took. We have
$$\begin{align}\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2}
&=\frac{1}{4}\sum_{n=0}^\infty \bigg(\frac{1}{2n-1}-\frac{1}{2n+1}\bigg)^2\\
&=\frac{1}{4}\sum_{n=0}^\infty \frac{1}{(2n-1)^2}+\frac{1}{(2n+1)^2}-\frac{2}{(2n-1)(2n+1)}
\end{align}$$
Now, using the fact that
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{3}{4}\zeta(2)=\frac{\pi^2}{8}$$
we have
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2}=\frac{3\zeta(2)+2}{8}-\frac{1}{2}\sum_{n=0}^\infty \frac{1}{(2n-1)(2n+1)}$$
Now we may use telescoping to evaluate the latter series as
$$\frac{1}{2}\sum_{n=0}^\infty \frac{1}{(2n-1)(2n+1)}=\frac{1}{4}\sum_{n=0}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}=-\frac{1}{4}$$
so that we have
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2}=\frac{3\zeta(2)+4}{8}=\frac{1}{2}+\frac{\pi^2}{16}$$
