Jun 12, 2021 - 22:34

An infinite series problem.

By Vamshi Jandhyala in mathematics

June 12, 2021

Problem

Solution

From here, we have

$$ \begin{align*} \coth x &= \sum_{n=-\infty}^{\infty} \frac{x}{x^{2}+\pi^{2}n^{2}} \\ &= \frac{1}{x} + 2 \sum_{n=1}^{\infty} \frac{x}{x^{2}+\pi^{2}n^{2}} \end{align*} $$

Therefore,

$$ \begin{aligned} \coth{\pi} &= \frac{1}{\pi} + \frac{2}{\pi} \sum_{n=1}^\infty \frac{1}{1+ n^2} \\
\frac{1}{1^2+1} + \frac{1}{2^2+1} + \cdots &= \frac{\pi\coth{\pi} -1}{2} \\
&= \frac{1}{2}\left(\pi \frac{e^{\pi} + e^{-\pi}} {e^{\pi}- e^{-\pi}} -1 \right) \end{aligned} $$

We also have

$$ \begin{aligned} e^0 &= \frac{\sinh{\pi}}{\pi} \left( 1 + 2 \sum_{n=1}^\infty \frac{(-1)^n}{1+ n^2} \right) \\
\implies \frac{1}{1^2+1} - \frac{1}{2^2+1} + \cdots &= \frac{1}{2} - \frac{\pi}{e^{\pi} - e^{-\pi}} \end{aligned} $$