<s>[INST] Evaluation of $\sum^{\infty} _{n=1} \arctan\left(\frac{4n}{n^4-2n^2+2}\right)$ 
Evaluate $\displaystyle\sum^{\infty} _{n=1} \arctan\left(\frac{4n}{n^4-2n^2+2}\right).$

I know we know to convert it in the of $\arctan\left(\frac{a-b}{1+ab}\right)$ but I am not able to do so here.
Could someone give me some hint?
 [/INST] Hint. One may write, for $n\ge2$,
$$
\frac{4n}{n^4-2n^2+2}=\frac{4n}{(n^2-1)^2+1}=\frac{\frac{4n}{(n^2-1)^2}}{1+\frac1{(n^2-1)^2}}=\frac{\frac1{(n-1)^2}-\frac1{(n+1)^2}}{1+\frac1{(n-1)^2(n+1)^2}}
$$ giving here
$$
\arctan\frac{4n}{n^4-2n^2+2}=\arctan\frac1{(n-1)^2}-\arctan\frac1{(n+1)^2}.
$$
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