# Evaluate $\sum_{n=1}^{\infty}\arctan (\frac{2}{n^2})$ [duplicate]

Question is to Evaluate $$\sum_{n=1}^{\infty}\arctan (\frac{2}{n^2})$$

What I have done so far is i tried taking sequence of patial sums.

$S_1=\arctan (\frac{2}{1})$

$S_2=\arctan (\frac{2}{1})+\arctan (\frac{2}{2^2})=\arctan 2+\arctan (\frac{1}{2})$

$\tan S_2=\tan(\arctan 2+\arctan (\frac{1}{2}))=\frac{***}{1-2.\frac{1}{2}}=\frac{***}{0}\Rightarrow S_2 =\frac{\pi}{2}$

Now there is a problem for $S_3$

$S_3=\arctan (\frac{2}{1})+\arctan (\frac{2}{2^2})+\arctan (\frac{2}{2^3})=\frac{\pi}{2}++\arctan (\frac{2}{2^3})$

I can not apply $\tan$ function on both sides as this would again give me :

$\tan S_3=\tan(\frac{\pi}{2}+\arctan (\frac{2}{2^3}))=\frac{\infty+**}{1-\infty**}$ which does not makes much sense.

I would be thankful if some one can help me with this.

Thank you.

P.S : I would be thankful if some one can suggest me another way to solve this than that of the originally posted solution in quoted duplicate...