Evaluation of $\sum^{\infty} _{n=1} \arctan\left(\frac{4n}{n^4-2n^2+2}\right)$

Question

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?

Answer 1

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 $$ \arctan\frac{4n}{n^4-2n^2+2}=\arctan\frac1{(n-1)^2}-\arctan\frac1{(n+1)^2}. $$

Answer 2

HINT:

$$\dfrac{4n}{n^4-2n^2+2}=\dfrac{(n+1)^2-(n-1)^2}{1+(n^2-1)^2}$$

$$\implies\arctan\dfrac{4n}{n^4-2n^2+2}=\arctan\{(n+1)^2\}-\arctan\{(n-1)^2\}=f(n+1)-f(n-1)$$

where $f(m)=\arctan(m^2)$