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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.

Sangchul Lee
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Ananthakrishnan
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    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 07:17
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    Related : https://math.stackexchange.com/questions/144944/finding-tan-t-if-t-sum-tan-11-2t2, https://math.stackexchange.com/questions/415512/is-s-sum-r-1-infty-tan-1-frac2r2r2r4-finite https://math.stackexchange.com/questions/617032/calculation-of-int-01-tan-11-xx2dx – lab bhattacharjee Sep 13 '17 at 07:57
  • Also an elementary approach is given here. It is works from the vein that @SangchulLee pointed out. – Mason Dec 27 '18 at 15:50

1 Answers1

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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.

Angina Seng
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  • Briefly, how did you get the first step? :( – Joao Noch Sep 13 '17 at 07:30
  • @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 07:50