TITLE: Series Question: $\sum_{n=1}^{\infty}\frac{1}{16n^2-1}$
QUESTION [6 upvotes]: How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{1}{16n^2-1}$$
I tried to use partial fraction
$$\frac{1}{16n^2-1}=\frac{1}{(4n-1)(4n+1)}=\frac{1}{2}\left[\frac{1}{4n-1}-\frac{1}{4n+1}\right]$$
I thought it will form telescoping series, but it is not. Any help would be appreciated. Thanks in advance.

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{n = 1}^{\infty}{1 \over 16n^{2} - 1}:\ {\large ?}}$

Even though the result is quite trivial, I was curious about the factor $\ds{\pi \over 4}$ which is equal to
  $\ds{\arctan\pars{1} = \int_{0}^{1}{\dd x \over x^{2} + 1}}$. So, this integral should be involved in the calculation. That's what I want to show here which adds another way to arrive to the result.

\begin{align}
\color{#66f}{\large\sum_{n = 1}^{\infty}{1 \over 16n^{2} - 1}}&
=\half\pars{{1 \over 4n - 1} - {1 \over 4n + 1}}
=-\,{1 \over 8}\sum_{n = 1}^{\infty}\sum_{\sigma = \pm}{\sigma \over n + \sigma/4}
\\[3mm]&=-\,{1 \over 8}\sum_{n = 1}^{\infty}\sum_{\sigma = \pm}\sigma
\int_{0}^{1}t^{n + \sigma/4 - 1}\,\dd t
=-\,{1 \over 8}\sum_{\sigma = \pm}\sigma\int_{0}^{1}
\sum_{n = 1}^{\infty}t^{n + \sigma/4 - 1}\,\dd t
\\[3mm]&=-\,{1 \over 8}\sum_{\sigma = \pm}\sigma
\int_{0}^{1}{t^{\sigma/4} \over 1 - t}\,\dd t
={1 \over 8}\int_{0}^{1}{1 - t^{1/2} \over t^{1/4}\pars{1 - t}}\,\dd t
\\[3mm]&={1 \over 8}\
\overbrace{\int_{0}^{1}{\dd t \over t^{1/4}\pars{1 + t^{1/2}}}}
^{\ds{\mbox{Set}\ t = x^{4}}}\
=\ \half\int_{0}^{1}{x^{2} \over 1 + x^{2}}\,\dd x
\\[3mm]&=\half\pars{\int_{0}^{1}\dd x - \int_{0}^{1}{\dd x \over 1 + x^{2}}}
=\half\bracks{1 - \arctan\pars{1}}
=\color{#66f}{\large\half\pars{1 - {\pi \over 4}}}
\end{align}