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?

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$$\frac14+\frac14+\frac{\pi^2}{16}=\frac12+\frac{\pi^2}{16}.$$
