Different approaches for evaluating a series I was playing around with the Fourier series of $\cos x , \; x \in (0, \frac{\pi}{2})$ and a series came up. First of all the fourier series of $\cos x$ at the given interval is:
$$\cos x =\frac{8}{\pi} \sum_{n=1}^{\infty} \frac{n \sin 2nx}{4n^2-1}$$
Calculations are pretty straightforward. By applying Parseval we get that:
$$\sum_{n=1}^{\infty} \frac{n^2}{(4n^2-1)^2}= \frac{\pi^2}{64}$$
What other approaches can we use in order to evaluate this series? Perhaps partial decomposition? One notes that:
$$\frac{n^2}{\left ( 4n^2-1 \right )^2}= \frac{1}{16}\left [ \frac{1}{\left ( 2n+1 \right )^2}- \frac{1}{(2n-1)^2} \right ]+ \frac{1}{16}\left [ \frac{1}{2n-1} - \frac{1}{2n+1} \right ]$$
Hence:
$$\sum_{n=1}^{\infty}\frac{n^2}{\left ( 4n^2-1 \right )^2}= \frac{1}{16}\sum_{n=1}^{\infty}\left [ \frac{1}{\left ( 2n+1 \right )^2}- \frac{1}{(2n-1)^2} \right ]+ \frac{1}{16}\sum_{n=1}^{\infty}\left [ \frac{1}{2n-1} - \frac{1}{2n+1} \right ]$$
However, we all know where the first two sums evaluate to. They are quite famous:
$\displaystyle  {\color{gray} \bullet} \;\;\; \sum_{n=1}^{\infty}\frac{1}{(2n+1)^2}= \frac{\pi^2}{8}- 1$
$\displaystyle {\color{gray} \bullet} \;\;\; \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}= \frac{\pi^2}{8}$
Hence the first bracket is evaluated to $-\frac{1}{16}$. The second bracket, is easily noted that it can be expressed in terms of digamma. However, I am having a difficult time converting to the digamma. 
Any help on that point would be nice. Also I am interest in other approaches, such as contour integration. 
 A: There's a slight mistake in your decomposition; it should be:
$$\frac{n^2}{\left ( 4n^2-1 \right )^2}= \frac{1}{16}\left [ \frac{1}{\left ( 2n+1 \right )^2} \color{red}{+} \frac{1}{(2n-1)^2} \right ]+ \frac{1}{16}\left [ \frac{1}{2n-1} - \frac{1}{2n+1} \right ]$$
Note that:
$$\sum_{k=1}^n \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) = -\sum_{k=1}^n (a_{k} - a_{k-1}) = -(a_n - a_0) = a_0 - a_n = 1 - \frac{1}{2n+1} \to 1$$
Where $a_k = \frac{1}{2k+1}, k=0,1,2,...$
A: I would define (valid when $x>1$) $$S(x)=\sum_{n=1}^{\infty} \frac1{x^2n^2-1}=\frac1{x^2}\sum_{n=1}^{\infty}\frac{1}{n^2-\frac1{x^2}}\\
=\frac12-\frac{\pi}{2x}\cot(\frac{\pi}{x})$$
This follows from logarithmically differentiating the Weierstrass product form of th sine function $$\frac{\sin(\pi z)}{\pi z}=\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right).$$
Differentiation yields $$S'(x)=\sum_{n=1}^{\infty} \frac{-2xn^2}{(x^2n^2-1)^2}=\frac{\pi}{2x^2}\cot(\frac{\pi}{x})-\frac{\pi^2}{2x^3}\csc^2(\frac{\pi}{x})$$ 
and letting $x=2$ gives the desired value.
