TITLE: How to get the sum of the series $\sum _{n=1}^{\infty } \frac{1}{\left(4 n^2-1\right)^2}$?
QUESTION [7 upvotes]: So, in order to obtain the required answer, I tried to apply some Taylor expansions, which led me to nowhere actually. 
After a while I tried to use the summation theorem 
$\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}}$ at $f\left(z\right)$'s poles.
Residue at $z=-\frac{1}{2}$ equals to residue at $z=\frac{1}{2}$, both of them are $-\frac{\pi ^2}{16}$. 
What I got seems to be not the correct answer after all:
$-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}} = -\frac{\pi ^2}{8}$, so that $\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{8}$
Since the $f\left(z\right)$ is even, I get $\sum_{n=0}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{16}$
However, the initial task was to find the sum of $\sum_{n=1}^{+\infty}{f\left(n\right)}$. I supposed that $\sum_{n=1}^{+\infty}{f\left(n\right)}=\sum_{n=0}^{+\infty}{f\left(n\right)}-f\left(0\right)=\frac{\pi ^2}{16}-1$, which is incorrect, as according to Mathematica, I should get $\frac{\pi ^2}{16}-\frac{1}{2}$. I don't know where is my mistake at this point. Perhaps, the whole approach is not well executed.

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{n = 1}^{\infty}{1 \over \pars{4n^{2} - 1}^{2}} & =
\left.{1 \over 16}\,\partiald{}{a}\sum_{n = 1}^{\infty}{1 \over n^{2} - a^{2}}
\right\vert_{\ a\ =\ 1/2} =
\left.{1 \over 16}\,\partiald{}{a}\sum_{n = 0}^{\infty}\pars{%
{1 \over n + 1 - a} - {1 \over n + 1 + a}}{1 \over 2a}
\right\vert_{\ a\ =\ 1/2}
\\[5mm] & =
\left.{1 \over 32}\,\partiald{}{a}\pars{H_{a} - H_{-a} \over a}
\right\vert_{\ a\ =\ 1/2}\qquad\qquad\quad\pars{~H_{z}:\ Harmonic\ Number~}
\\[5mm] & =
\left.{1 \over 32}\,\partiald{}{a}\pars{H_{a - 1} + 1/a - H_{-a} \over a}
\right\vert_{\ a\ =\ 1/2}\qquad\pars{~H_{z}\ Recursive\ Property~}
\\[5mm] & =
\left.{1 \over 32}\,\partiald{}{a}\pars{{1 \over a^{2}} -
{H_{-a} - H_{a - 1} \over a}}\right\vert_{\ a\ =\ 1/2}
\\[5mm] & =
\left.{1 \over 32}\,\partiald{}{a}\pars{{1 \over a^{2}} -
{\pi\cot\pars{\pi a} \over a}}\right\vert_{\ a\ =\ 1/2}
\qquad\pars{~H_{z}\ Euler\ Reflection\ Formula~}
\\[5mm] & =
\left.{1 \over 32}\pars{-\,{2 \over a^{3}} +
{\pi\cot\pars{\pi a} \over a^{2}} +
{\pi^{2}\csc^{2}\pars{\pi a} \over a}}\right\vert_{\ a\ =\ 1/2} =
\bbx{\pi^{2} - 8 \over 16} \approx 0.1169
\end{align}