A telescopic (!) series

Prove that

$\sum_{n=-\infty}^{\infty} \arctan \frac{\sinh 1}{\cosh 2n} = \frac{\pi}{2}$

Solution

We have successively:

$\begin{align*} \sum_{n=-\infty}^{\infty} \arctan \frac{\sinh 1}{\cosh 2n} &= \arctan \sinh 1 + 2 \sum_{n=1}^{\infty} \arctan \frac{\sinh 1}{\cosh 2n} \\ &= \arctan \frac{e-e^{-1}}{2} + 2 \sum_{n=1}^{\infty}\arctan \frac{e-e^{-1}}{e^{2n} + e^{-2n}} \\ &= \arctan \frac{e - e^{-1}}{1 + e \cdot e^{-1}} + 2 \sum_{n=1}^{\infty} \arctan \frac{e^{-(2n-1) - e^{-(2n+1)}}}{1 + e^{-4n}} \\ &= \arctan \frac{e - e^{-1}}{1 + e \cdot e^{-1}} + 2 \sum_{n=1}^{\infty} \arctan \frac{e^{-(2n-1) - e^{-(2n+1)}}}{1 + e^{-(2n-1)} e^{-(2n+1)}}\\ &= \left ( \arctan e - \arctan e^{-1} \right ) + \\ &\quad \quad \quad + 2 \sum_{n=1}^{\infty} \left ( \arctan e^{-(2n-1)} - \arctan e^{-(2n+1)} \right ) \\ &= \left ( \arctan e - \arctan e^{-1} \right ) + 2 \arctan e^{-1} \\ &= \arctan e + \arctan e^{-1} \\ &= \frac{\pi}{2} \end{align*}$

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A telescopic (!) series

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