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?

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
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2}=\frac{3\zeta(2)+4}{8}=\frac{1}{2}+\frac{\pi^2}{16}$$
