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.


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.

  • $\begingroup$ Briefly, how did you get the first step? :( $\endgroup$ – Joao Noch Sep 13 '17 at 7:30
  • $\begingroup$ @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. $\endgroup$ – Sangchul Lee Sep 13 '17 at 7:50

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