<s>[INST] 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.
 [/INST] 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
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