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Prove that $$\sum_{n\in\mathbb{Z}}\arctan\left(\frac{\sinh(1)}{\cosh(2n)}\right)=\frac{\pi}{2}$$

Writing $$\dfrac{\sinh(1)}{\cosh(2n)}=\dfrac{e^{1}-e^{-1}}{e^{2n}+e^{-2n}}$$

I tried to use the identity $$\arctan\left(\frac{a_1}{a_2}\right)+\arctan\left(\frac{b_1}{b_2}\right)=\arctan\left(\frac{a_1b_2+ a_2b_1}{a_2b_2-a_1b_1}\right)$$ with a suitable choice of $a_1,a_2,b_1,b_2$ but I haven't been able to find a telescopic sum.

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    Do you real mean $a_b$ or is it $a_2$? – 2017-02-07
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    Yes, thanks for your correction. – 2017-02-07
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    $$ \sum_j \arctan x_j$$ $$ = \arctan \frac{\sum_j x_j - \sum_{j,k,\ell} x_j x_k x_\ell + \sum\text{products of five} - \sum\text{products of seven} + \cdots }{1 - \sum_{j,k} x_j x_k + \sum\text{products of four} - \sum\text{products of six} + \cdots}$$ I don't know if this will get you anywhere or not. – 2017-02-07
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    @MichaelHardy: the trick is just to consider the unsual function $\arctan\tanh$. This problem appeared in the *Monthly* and in my course notes (https://drive.google.com/file/d/0BxKdOVsjsuEwcGlSbzgzZVdYZkk/view) almost at the same time, independently :) – 2017-02-07

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That is a telescopic sum in disguise. We may notice that:

$$\arctan\tanh(n+1)-\arctan\tanh(n-1) = \arctan\left(\frac{\tanh(n+1)-\tanh(n-1)}{1+\tanh(n-1)\tanh(n+1)}\right) $$ equals $\arctan\left(\frac{\sinh(2)}{\cosh(2n)}\right)$ and: $$ \arctan\left(\frac{\sinh(1)}{\cosh(2n)}\right) = \arctan\tanh\left(n+\frac{1}{2}\right)-\arctan\tanh\left(n-\frac{1}{2}\right). $$ You may easily draw your conclusions now.

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    nice observation, thanks @Jack D'Aurizio. – 2017-02-07