How to compute $\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)$ I find this problem on facebook group.
$$\mbox{Is it possible to find exact value of}\quad
\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)\ {\large ?}.
$$
I think this is not telescope sum. And Wolfram Alpha can not find it.
Thank in advances.
 A: @Jack DAurizio answer is nice,and I have two solution  for this
\begin{align*}\arctan{\dfrac{1}{2n^2}}&=\arctan{\dfrac{2}{4n^2-1+1}}=\arctan{\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}}\\
&=\arctan{(2n+1)}-\arctan{(2n-1)}
\end{align*}
solution 2:
\begin{align*}\arctan{\dfrac{1}{2n^2}}&=\arctan{\dfrac{n^2-(n^2-1)}{(n^2-n)+(n^2+n)}=\arctan{\dfrac{\dfrac{n}{n-1}-\dfrac{n+1}{n}}{1+\dfrac{n}{n-1}\cdot\dfrac{n+1}{n}}}}\\
&=\arctan{\dfrac{n}{n-1}}-\arctan{\dfrac{n+1}{n}}
\end{align*}
and for $$\sum_{n=1}^{\infty}\arctan{\dfrac{1}{n^2}}=\arctan{\dfrac{\tan{\frac{\pi}{\sqrt{2}}}-\tanh{\frac{\pi}{\sqrt{2}}}}{\tan{\frac{\pi}{\sqrt{2}}}+\tanh{\frac{\pi}{\sqrt{2}}}}}$$
see this (AMM E3375) post: prove this $\sum_{n=1}^{\infty}\arctan{\left(\dfrac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)\cdots\right)}$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large
\sum_{n\ =\ 1}^{\infty}\arctan\pars{3n^{2} \over 2n^{4} - 1}}
=\sum_{n\ =\ 1}^{\infty}\bracks{
\arctan\pars{1 \over n^{2}} + \arctan\pars{1 \over 2n^{2}}}
\\[5mm]&=\sum_{k\ =\ 1}^{2}
\dsc{\sum_{n\ =\ 1}^{\infty}\arctan\pars{1 \over kn^{2}}}\tag{1}
\end{align}

\begin{align}&\dsc{\sum_{n\ =\ 1}^{\infty}\arctan\pars{1 \over kn^{2}}}
=\sum_{n\ =\ 1}^{\infty}\int_{0}^{1}{kn^{2} \over x^{2} + k^{2}n^{4}}\,\dd x
={1 \over k}\Re\int_{0}^{1}\sum_{n\ =\ 1}^{\infty}{1 \over n^{2} + x\ic/k}\,\dd x
\\[5mm]&={1 \over k}\,\Re\int_{0}^{1}\sum_{n\ =\ 0}^{\infty}
{1 \over \pars{n + 1 + \root{x\ic/k}}\pars{n + 1 - \root{x\ic/k}}}\,\dd x
\\[5mm]&={1 \over k}\,\Re\int_{0}^{1}
{\Psi\pars{1 + \root{x\ic/k}} - \Psi\pars{1 - \root{x\ic/k}} \over 2\root{x\ic/k}}\,\dd x
\end{align}
  where $\ds{\Psi}$ is the Digamma Function.

With the change $\ds{\root{{x \over k}\,\ic}=t\ \imp\ x=-kt^{2}\,\ic}$:
\begin{align}&\dsc{\sum_{n\ =\ 1}^{\infty}\arctan\pars{1 \over kn^{2}}}
={1 \over k}\,\Re\int_{0}^{\root{\ic/k}}
{\Psi\pars{1 + t} - \Psi\pars{1 - t} \over 2t}\,\pars{-2kt\,\ic\,\dd t}
\\[5mm]&=\Im\int_{0}^{\root{\ic/k}}
\bracks{\Psi\pars{1 + t} - \Psi\pars{1 - t}}\,\dd t
=\Im\int_{0}^{\root{\ic/k}}
\braces{{1 \over t} - \bracks{\Psi\pars{1 - t} - \Psi\pars{t}}}\,\dd t
\\[5mm]&=\Im\int_{0}^{\root{\ic/k}}
\bracks{{1 \over t} - \pi\cot\pars{\pi t}}\,\dd t
=\Im\bracks{
\ln\pars{\root{\ic \over k}} - \ln\pars{\sin\pars{\pi\root{\ic \over k}}}}
\end{align}

However
  \begin{align}
&\color{#00f}{\Im\ln\pars{\root{\ic \over k}}}=\Im\ln\pars{1 + \ic \over \root{2k}}
=\color{#00f}{{\pi \over 4}}
\\[1cm]
&\color{#00f}{\Im\ln\pars{\sin\pars{\pi\root{\ic \over k}}}}
=\Im\ln\pars{\sin\pars{{\pi \over \root{2k}} + {\pi \over \root{2k}}\ic}}
\\[5mm]&=\Im\ln\pars{\sin\pars{\pi \over \root{2k}}\cosh\pars{\pi \over \root{2k}}
+\cos\pars{\pi \over \root{2k}}\sinh\pars{\pi \over \root{2k}}\ic}
\\[5mm]&=\color{#00f}{
\arctan\pars{\cot\pars{\pi \over \root{2k}}\tanh\pars{\pi \over \root{2k}}}}
\end{align}
  such that
  $$
\dsc{\sum_{n\ =\ 1}^{\infty}\arctan\pars{1 \over kn^{2}}}\
{\large =}\ \dsc{{\pi \over 4}
-\arctan\pars{\cot\pars{\pi \over \root{2k}}\tanh\pars{\pi \over \root{2k}}}}
$$

With expression $\pars{1}$:
\begin{align}&\color{#66f}{\large
\sum_{n\ =\ 1}^{\infty}\arctan\pars{3n^{2} \over 2n^{4} - 1}}
=\color{#66f}{\large{\pi \over 2}
-\arctan\pars{\cot\pars{\pi \over \root{2}}\tanh\pars{\pi \over \root{2}}}}
\\[5mm]&\approx{\tt 2.21013994}
\end{align}
A: Let $$\tan x=\frac{1}{n^2} \text{and } \tan y=\frac{1}{2n^2}$$  so we have that $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{\frac{1}{n^2}+\frac{1}{2n^2}}{1-\left(\frac{1}{n^2}\frac{1}{2n^2}\right)}=\frac{3n^2}{2n^4-1}$$ hence we have $$x+y=\arctan\frac{\frac{1}{n^2}+\frac{1}{2n^2}}{1-\left(\frac{1}{n^2}\frac{1}{2n^2}\right)}=\arctan\frac{1}{n^2}+\arctan\frac{1}{2n^2}.$$ From here I think you can do
A: Since:
$$\arctan\frac{3n^2}{2n^4-1}=\arctan\frac{1}{n^2}+\arctan\frac{1}{2n^2}$$
then:
$$\sum_{n=1}^{+\infty}\arctan\frac{3n^2}{2n^4-1} = \arg\prod_{n=1}^{+\infty}\left(1+\frac{i}{n^2}\right)+\arg\prod_{n=1}^{+\infty}\left(1+\frac{i}{2n^2}\right).\tag{1} $$
Since, by the Weierstrass product for the $\sinh$ function,
$$\frac{\sinh(\pi z)}{\pi z}=\prod_{n=1}^{+\infty}\left(1+\frac{z^2}{n^2}\right),$$
and by telescopic property we have:
$$\sum_{n=1}^{+\infty}\arctan\frac{1}{2n^2}=\frac{\pi}{4},$$
it happens that:
$$\begin{eqnarray*}\sum_{n=1}^{+\infty}\arctan\frac{3n^2}{2n^4-1} &=& \frac{\pi}{4}+\arg\frac{\sinh(\pi e^{i\pi/4})}{\pi e^{i\pi/4}}=\arg\sinh\left(\pi e^{i\pi/4}\right)\\&=&\arg\sinh\left(\frac{\pi}{\sqrt{2}}(1+i)\right)\\&=&\color{red}{\pi+\arctan\left(\tan\frac{\pi}{\sqrt{2}}\coth\frac{\pi}{\sqrt{2}}\right)}.\tag{2}\end{eqnarray*}$$
