Find the summation of $\sum_{n=1}^{\infty }\frac{1}{(2n)^2-1}$ I need to find the summation of the above series in closed form.
 A: First observe that $$
\frac{1}{n^2-1}=\frac{1}{(n+1)(n-1)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)
$$ and conclude by telescoping terms.
You end up with
$$
\sum_{n=2}^{N}\frac{1}{n^2-1}=\frac{1}{2}\left(1-\frac{1}{N+1}\right)
$$ and then let $N$ tend to $+\infty$ to obtain $\displaystyle  \frac12$ as the sum of the series $\displaystyle \sum_{n=2}^{\infty}\frac{1}{n^2-1}$.
Now it is easy to adapt the previous steps to the series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n)^2-1}.\quad$  :)!
You may start with $$ \frac{1}{(2n)^2-1}=\frac{1}{(2n+1)(2n-1)}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$
and then...
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#66f}{\large\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2} - 1}}
={1 \over 4}\sum_{n\ =\ 0}^{\infty}{1 \over \pars{n + 3/2}\pars{n + 1/2}}
={1 \over 4}\,{\Psi\pars{3/2} - \Psi\pars{1/2} \over 3/2 - 1/2}
\\[3mm]&={1 \over 4}\,\bracks{\Psi\pars{3 \over 2} - \Psi\pars{\half}}\tag{1}
\end{align}
where $\ds{\Psi\pars{z}}$ is the
Digamma Function $\color{#000}{\bf 6.3.1}$.

With the property $\ds{\Psi\pars{z} = \Psi\pars{z + 1} - {1 \over z}}$ we'll have 
  $$
\Psi\pars{3 \over 2} - \Psi\pars{\half}
=\Psi\pars{3 \over 2} - \bracks{\Psi\pars{\half + 1} - {1 \over 1/2}}
=2
$$

By replacing in $\pars{1}$:
$$
\color{#66f}{\large\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2} - 1}}
={1 \over 4}\times2=\color{#66f}{\Large\half}
$$
