I was watching a Numberphile video (on how $\tan^{-1}{1} + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \frac{\pi}{2}$) and I thought about whether the series $$\sum_{k = 1}^{\infty}{\tan^{-1}}{\frac{1}{k}}$$


Turns out, it doesn't. However, it also turns out that $$\sum_{k = 1}^{\infty}{\tan^{-1}}{\frac{1}{k^2}}$$ does converge, to the expression $$\tan^{-1}{\left(\frac{ {1-\cot{\dfrac{\pi}{\sqrt{2}}} \tanh{\dfrac{\pi}{\sqrt{2}}}} }{ {1+\cot{\dfrac{\pi}{\sqrt{2}}} \tanh{\dfrac{\pi}{\sqrt{2}}}} }\right)}$$

Could someone help me understand how this comes about?

  • $\begingroup$ The resulting fraction is intriguingly similar to a summand of this question, though that may merely be coincidental. $\endgroup$ Commented Oct 6, 2014 at 15:20
  • 3
    $\begingroup$ $$\tan^{-1}(t) = \frac{1}{2i} \log\left(\frac{1+it}{1-it}\right) \quad\text{ AND }\quad\frac{\sinh\pi x}{\pi x} = \prod_{k=1}^\infty \left(1+\frac{x^2}{k^2}\right)$$ $\endgroup$ Commented Oct 6, 2014 at 15:55
  • $\begingroup$ A related question. $\endgroup$
    – Lucian
    Commented Oct 6, 2014 at 16:14
  • $\begingroup$ It turns out a similar question has been asked before and has good answers. I wonder whether we should close this as an abstract duplicate? $\endgroup$ Commented Oct 6, 2014 at 16:41

1 Answer 1


Use the logarithmic form $\arctan x=\frac1{2i}\ln\frac{1+ix}{1-ix}$ to rewrite the sum as $$S=\frac1{2i}\ln\prod_{k=1}^{\infty}\frac{1+\frac{i}{k^2}}{1-\frac{i}{k^2}}$$ and combine the result with Euler's infinite product $$\prod_{k=1}^{\infty}\left(1+\frac{a^2}{k^2}\right)=\frac{\sinh \pi a}{\pi a}.$$ This gives $$S=\frac{1}{2i}\ln\left(e^{-i\pi/2}\frac{\sinh \pi e^{i\pi/4}}{\sinh \pi e^{-i\pi/4}}\right)=\frac{1}{2i}\ln\left(\frac{\sinh \frac{\pi}{\sqrt 2}\cos\frac{\pi}{\sqrt 2} +i\cosh \frac{\pi}{\sqrt 2}\cos\frac{\pi}{\sqrt 2}}{\cosh \frac{\pi}{\sqrt 2}\sin\frac{\pi}{\sqrt 2} +i\sinh \frac{\pi}{\sqrt 2}\cos\frac{\pi}{\sqrt 2}}\right),$$ which can be straightforwardly reduced to the quoted result.


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