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.
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.