Summation containing $\pi$ gives an expression containing $e$

Question

This is an interesting relation between $\pi$ and $e$.

$$\sum_{k=1}^{\infty }\frac{1}{(k\pi)^2+1}=\frac{1}{\pi^2+1}+\frac{1}{4\pi^2+1}+\frac{1}{9\pi^2+1}+\frac{1}{16\pi^2+1}+\dots=\frac{1}{e^2-1}$$

I tried to prove it using Fourier series expansion, but I could not. Any hint will be appreciated.

Poisson summation formula? – Angina Seng Jul 04 '19 at 06:04 — Angina Seng, Jul 04 '19 at 06:04

robjohn · Accepted Answer · 2019-07-04T14:06:10.137

6

Using the result that $$ \mathrm{PV}\sum_{k\in\mathbb{Z}}\frac1{x+k}=\pi\cot{\pi x} $$ we get $$ \begin{align} \sum_{k=1}^\infty\frac1{1+k^2\pi^2} &=\frac{i}{2\pi}\mathrm{PV}\sum_{k\in\mathbb{Z}}\frac1{i/\pi+k}-\frac12\\ &=\frac{i}{2\pi}\pi\cot\left(\pi\frac{i}\pi\right)-\frac12\\[3pt] &=\frac12\coth(1)-\frac12\\[3pt] &=\frac1{e^2-1} \end{align} $$

edited Jul 04 '19 at 14:06

answered Jul 04 '19 at 05:07

robjohn

345,667

Two different approaches—I love it! – gen-ℤ ready to perish Jul 04 '19 at 05:08
That first formula holds only if you interpret the left side as the "principal value" of the sum. – Gerry Myerson Jul 04 '19 at 10:17
@GerryMyerson: indeed (and this is mentioned in the cited link). However, for conditionally convergent series, we do assume a certain order with $\sum\limits_{n=0}^\infty$. So perhaps it would be better to write $\sum\limits_{|n|=0}^\infty$ instead of $\sum\limits_{n\in\mathbb{Z}}$. – robjohn Jul 04 '19 at 12:30
Yes, it's mentioned in the link. But it should be mentioned in your answer. – Gerry Myerson Jul 04 '19 at 12:59
1

And it was as soon as I posted my comment (as per my own recommendation). Oops, I missed that there were two instances. – robjohn Jul 04 '19 at 14:01

score 4 · Answer 2 · answered Jul 04 '19 at 04:11

4

Hint: Use the expansion $$\coth x=\sum_{n=-\infty}^{\infty}\dfrac{x}{n^2\pi^2+x^2}$$

answered Jul 04 '19 at 04:11

Nosrati

29,995

OK, but how do you get that expansion? – Gerry Myerson Jul 04 '19 at 04:35
@GerryMyerson I suppose you just have to be familiar with it. Deriving that is a whole separate matter – gen-ℤ ready to perish Jul 04 '19 at 05:10
@GerryMyerson see https://en.wikipedia.org/wiki/Mittag-Leffler%27s_theorem this is Mittag-Leffler's theorem for coth. – Nosrati Jul 04 '19 at 05:24
haha :) why downvote? – Nosrati Jul 04 '19 at 05:27
@Nosrati I didn't downvote, but wanted to say that Mittag-Leffler is not enough to prove this identity. There are infinitely many meromorphic functions with the same expansions at every pole. It is far from clear that the sum you write gives $\coth$ – Wojowu Jul 04 '19 at 07:13
@Nosrati It is. I am just adressing your comment regarding how to get the expansion. – Wojowu Jul 04 '19 at 07:20
@Wojowu thank you :) – Nosrati Jul 04 '19 at 07:21

Summation containing $\pi$ gives an expression containing $e$

2 Answers2