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?

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:
$$ \sum_{n\geq 1}\frac{1}{(4n^2-1)^2} = \color{red}{\frac{\pi^2}{16}-\frac{1}{2}}.$$
