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.
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: $$\begin{eqnarray*}\sum_{n=1}^{+\infty}\arctan\frac{3n^2}{2n^4-1} &=& \frac{\pi}{4}+\arg\frac{\sinh(\pi e^{i\pi/4})}{\pi e^{i\pi/4}}=\arg\sinh\left(\pi e^{i\pi/4}\right)\\&=&\arg\sinh\left(\frac{\pi}{\sqrt{2}}(1+i)\right)\\&=&\color{red}{\pi+\arctan\left(\tan\frac{\pi}{\sqrt{2}}\coth\frac{\pi}{\sqrt{2}}\right)}.\tag{2}\end{eqnarray*}$$
@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 $\sum_{n=1}^{\infty}\arctan{\left(\dfrac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)\cdots\right)}$
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}
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
