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This question already has an answer here:

How to derive the formula $$\sum_{n=1}^{\infty}\tan^{-1}\left(\frac{x^{2}}{n^{2}}\right)=\tan^{-1}\left[\frac{\tan\left(\frac{x\pi}{\sqrt{2}}\right)-\tanh\left(\frac{x\pi}{\sqrt{2}}\right)}{\tan\left(\frac{x\pi}{\sqrt{2}}\right)+\tanh\left(\frac{x\pi}{\sqrt{2}}\right)}\right]$$

Is the formula correct? How do I derive the above formula? I tried to apply the formula integral as the limit of sum formula.

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marked as duplicate by alexjo, MathOverview, Ben Sheller, Shailesh, Simply Beautiful Art Oct 7 '17 at 0:36

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    $\begingroup$ When you ask "is the formula correct?" do you have any reason to believe it is correct or incorrect? What have you tried so far? Some context would help $\endgroup$ – TomGrubb Oct 6 '17 at 16:07
  • $\begingroup$ Maybe index starts at $n=1$ $\endgroup$ – Raffaele Oct 6 '17 at 16:42
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I'd start that sum at $n=1$ if I were you.

Now, up to multiples of $\pi$, $$\sum_{n=1}^\infty\tan^{-1}\frac{x^2}{n^2} $$ is the argument of the complex number $$\prod_{n=1}^\infty\left(1+\frac{ix^2}{n^2}\right).$$

But $$\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)=\frac{\sin\pi z}{\pi z}$$ so how about choosing $z$ to make $-z^2=ix^2$?

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  • $\begingroup$ This is the classical infinite product for the sine. See texts on complex variables for proofs. @N.Maneesh $\endgroup$ – Lord Shark the Unknown Oct 6 '17 at 17:10
  • $\begingroup$ Thank you. I will find it. Can you give hints how to derive the formula given in question from your steps? Can you please help me? $\endgroup$ – Unknown x Oct 6 '17 at 17:13

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