# Find the value of infinite series $\sum_{n=1}^{\infty} \tan^{-1}(2/n^2)$

Find the value of infinite series

$$\sum_{n=1}^{\infty} \tan^{-1}\left(\frac{2}{n^2} \right)$$

I tried to find sequence of partial sums and tried to find the limit of that sequence. But I didn't get the answer.

• If I remember correctly, this problem goes back to Ramanujan. And it can be computed using telescoping technique combined with the following identity: $$\tan^{-1}\left(\frac{1}{n-1}\right) - \tan^{-1}\left(\frac{1}{n+1}\right) = \arctan\left(\frac{2}{n^2}\right).$$ – Sangchul Lee Sep 13 '17 at 7:17
• – lab bhattacharjee Sep 13 '17 at 7:57
• Also an elementary approach is given here. It is works from the vein that @SangchulLee pointed out. – Mason Dec 27 '18 at 15:50

It's the argument of the complex number $$\prod_{n=1}^\infty\left(1+\frac{2i}{n^2}\right) =\prod_{n=1}^\infty\frac{(n-1+i)(n+1-i)}{n^2} =\prod_{n=1}^\infty\frac{(n-1+i)(n^2+2n)}{n^2(n+1+i)}.$$ This infinite product telescopes.
• @JoaoNoch, It follows from $\arctan(x) = \arg(1+ix)$. Perhaps the only non-trivial part is that the argument of the product above determines OP's sum only modulo $2\pi$. But an estimate $$\sum_{n=1}^{\infty} \arctan(2/n^2) \leq \frac{\pi}{2} + 2(\zeta(2)-1) < \pi$$ shows that the principal argument function is enough for our purpose. – Sangchul Lee Sep 13 '17 at 7:50