Value of this infinite sum I want to compute the value of
$$\sum_{n=1}^{\infty} \frac{1}{((2n)^2 - 1)^2}.$$
I have tried evaluating the first couple partial sums, but can't find any recurrence. I don't have too many tools in my toolbox to proceed... Can anyone see how I could go on here?
 A: Hint:
$$\frac{1}{((2n)^2 - 1)^2} = \frac{1}{4}\left( \frac{1}{2 n+1}+\frac{1}{(2 n+1)^2}-\frac{1}{2 n-1}+\frac{1}{(2 n-1)^2}\right)$$
Then, you can start with partial sums, breaking them in four parts and doing a change of indices on 2 of them to get something which would (hopefully — I didn't go further in the details) simplify.
Edit: In particular, provided I didn't do any mistake in the process, you can show that
$$
\sum_{n=1}^N \frac{1}{((2n)^2 - 1)^2} = \frac{1}{4(2N+1)} + \frac{1}{4(2N+1)^2} + \frac{1}{2}\sum_{n=1}^N \frac{1}{(2n+1)^2}$$
where the first two terms go to $0$, and the last one goes to $\frac{1}{2}\sum_{n=1}^\infty \frac{1}{(2n+1)^2}$, which can be computed exactly (knowing that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ and breaking the latter series' partial sum into even and odd $n$'s).
A: You may use residue theory.  For example, a convergent sum
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{Res}_{z=z_k} \pi \cot{\pi z} \, f(z)$$
where the $z_k$ are the non-integer poles of $f$.  In your case, $f(z)=1/(4 z^2-1)^2$, so that the sum
$$\sum_{n=-\infty}^{\infty} \frac{1}{(4 n^2-1)^2} = -\frac{\pi}{16} \left (\left [\frac{d}{dz} \frac{\cot{\pi z}}{(z+1/2)^2}  \right]_{z=1/2} + \left [\frac{d}{dz} \frac{\cot{\pi z}}{(z-1/2)^2}  \right]_{z=-1/2}\right )$$
which, when evaluated, is $\pi^2/8$.  However, this is not the sum desired; rather, it is over $1$ to $\infty$.  In this case, the sum becomes
$$\sum_{n=1}^{\infty} \frac{1}{(4 n^2-1)^2} = \frac12 \left( \frac{\pi^2}{8}-1\right) \approx 0.116850$$
A: Hint: Use partial fractions as Clement C. suggests. One part telescopes, and the other two parts  are close relatives of the familiar $\sum \frac{1}{n^2}$, which maybe can be taken as known to be $\frac{\pi^2}{6}$.
A detail:
$$1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots =\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\right) -\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\right).$$
