A curious identity involving $e$ and $\pi$

An infinite series problem.

By Vamshi Jandhyala in mathematics

April 2, 2020

Prove

$$ \begin{align*} \frac{1}{1+\pi^2} + \frac{1}{1+4\pi^2} + \cdots = \frac{1}{e^2-1} \end{align*} $$

Solution

We first derive the series expansion of $coth(x)$ using the Fourier transform.

The expansion can be derived from the complex Fourier series of the $2\pi$-periodic function

$$ \begin{align*} f(x) = e^{ax}, - \pi < x < \pi \end{align*} $$

By definition of the complex Fourier series,

$$ \begin{align*} e^{ax} = \lim_{N \to \infty}\sum_{n=-N}^{N} c_{n} e^{inx} \end{align*} $$

where

$$ \begin{align*} c_{n} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{ax} e^{-inx} \ dx &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{(a-in)x} \ dx \\ &= \frac{1}{2 \pi} \frac{e^{(a-in)x}}{a-in} \Big|^{\pi}_{-\pi} \\ &= \frac{1}{2 \pi} \frac{1}{a-in} \Big( e^{a \pi}e^{-in \pi} - e^{- a \pi}e^{i n \pi} \Big) \\ &= \frac{1}{\pi} \frac{(-1)^{n}}{a-in}\frac{e^{a \pi}-e^{- a \pi}}{2} \\ &= \frac{(-1)^{n}}{\pi} \frac{a+in}{a^{2}+n^{2}} \sinh a \pi \end{align*} $$

At $x=\pi$ (a point of discontinuity),

$$ \begin{align*} \frac{e^{a \pi}+e^{- a \pi}}{2} = \cosh a \pi = \lim_{N \to \infty}\sum_{n=-N}^{N} c_{n} (-1)^{n} = \frac{\sinh a \pi}{\pi} \lim_{N \to \infty} \sum_{n=-N}^{N} \frac{a+in}{a^{2}+n^{2}} \end{align*} $$

which implies

$$ \begin{align*} \coth a \pi &= \frac{1}{\pi} \lim_{N \to \infty}\sum_{n=-N}^{N} \frac{a+in}{a^{2}+n^{2}} \\ &= \frac{1}{\pi} \lim_{N \to \infty}\sum_{n=-N}^{N} \frac{a}{a^{2}+n^{2}} + \frac{i}{\pi} \lim_{N \to \infty} \sum_{n=-N}^{N} \frac{n}{a^{2}+n^{2}} \\ &= \frac{1}{\pi} \lim_{N \to \infty}\sum_{n=-N}^{N} \frac{a}{a^{2}+n^{2}}+ 0 \\ &= \frac{1}{\pi}\sum_{n=-\infty}^{\infty} \frac{a}{a^{2}+n^{2}}
\end{align*} $$

And replacing $a$ with $\frac{x}{\pi}$, we get

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

We now have for $x=1$,

$$ \begin{align*} \frac{e^2+1}{e^2-1} - 1 = 2\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} \\
\implies \frac{1}{e^2-1} = \sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} \end{align*} $$