Series Question: $\sum_{n=1}^{\infty}\frac{1}{16n^2-1}$ 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.
 A: Rewrite
\begin{align}
\sum_{n=1}^\infty\frac1{16n^2-1}&=\sum_{n=1}^\infty\frac1{(4n-1)(4n+1)}\\
&=\sum_{n=1}^\infty\frac12\left(\frac1{4n-1}-\frac1{4n+1}\right)\\
&=\frac12\left(\frac{1}{3}-\frac{1}5+\frac{1}7-\frac{1}9+\frac{1}{11}-\frac1{13}+\cdots\right).
\end{align}
Recall Leibniz series
$$
\frac\pi4=1-\frac13+\frac15-\frac17+\frac19-\frac1{11}+\cdots
$$
then
$$
\sum_{n=1}^\infty\frac1{16n^2-1}=\color{blue}{\frac12\left(1-\frac\pi4\right)}.
$$
A: Generally speaking, $~\displaystyle\sum_{n=-\infty}^\infty\dfrac{x^2}{x^2-n^2}=\pi x\cdot\cot(\pi x)$. In this case, $x=\dfrac14$ . This can be proven by 
taking the natural logarithm of Euler's infinite product expression for $\dfrac{\sin(\pi x)}{\pi x}$, and differentiating 
both sides. If the minus sign in the denominator would be changed to a plus sign, then the series 
would become $\pi x\cdot\coth(\pi x)$.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\sum_{n = 1}^{\infty}{1 \over 16n^{2} - 1}:\ {\large ?}}$

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}
