# Proving That ζ(2) = π2/6 With Your Eyes Closed

“Next time you go clubbing, ask around what’s the sum over k of 1 / k2” once said my math teacher. “You’ll see who’s worth partying with”.

Here’s what you need to know:

$\zeta(2) = \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$

While Euler was the first to prove the identity in 17401, a handful of other modern approaches have been proposed since then2:

1. Using a monotone convergence3
2. Using the Jacobian Matrix2
3. Using the power series for the inverse sine function4
4. Using the L2-completeness of the trigonometric functions2
5. Using the convergence of Fourier series2
6. Using uniform convergence of a power series on the real line2
7. Using an infinite product for the sine function1
8. Using the calculus of residues2
9. Using cotangente inequalities5
10. Using the MacLaurin expansion6
11. Using the reduction of integral7
12. Using the identity for the Fejér kernel8
13. Using Gregory’s formula and infinite limits9
14. Using the formula for the number of representations of a positive integer as a sum of four squares10

I was recently reminded of the Jacobian Matrix, let’s use this method for example.

Before beginning

We know that2

$\frac{3}{4} \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{m=1}^{\infty}\frac{1}{(2m)^2} = \sum_{r=0}^{\infty} \frac{1}{(2r + 1)^2},$

and that

$\sum_{r=0}^{\infty} \frac{1}{(2r + 1)^2} = \frac{\pi^2}{8}.$

We also note that

$\frac{1}{n^2} = \int_{0}^{1}\int_{0}^{1} x^{n-1} y^{n-1} dx\:dy,$

gives

$\sum_{n=1}^{\infty} \frac{1}{n^2} = \int_{0}^{1}\int_{0}^{1} \frac{dx\:dy}{1-xy}$

by monotonous convergence3.

Off we go

We start considering the monotonous convergence:

$\sum_{r=0}^{\infty} \frac{1}{(2r + 1)^2} = \int_{0}^{1}\int_{0}^{1} \frac{dx\:dy}{1-x^2y^2}.$

Let’s use the substitution

$(u, v) = \left( \tan^{-1} x \sqrt{\frac{1-y^2}{1-x^2}}, \tan^{-1} y \sqrt{\frac{1-x^2}{1-y^2}} \right),$

so that

$(x, y) = \left( \frac{\sin u}{\cos v}, \frac{\sin v}{\cos u} \right).$

The Jacobian matrix is:

\begin{align*} \frac{\partial(x, y)}{\partial(u, v)} &= \begin{vmatrix} \cos u / \cos v & \sin u \sin v / \cos^2 v\\ \sin u \sin v / \cos^2u & \cos v / \cos u \end{vmatrix} \\ &= 1- \frac{\sin^2 u \sin^2 v}{\cos^2 u \cos^2 v} \\ &= 1-x^2y^2. \end{align*}

We get:

$\frac{3}{4} \zeta(2) = \int\int_{A} du\:dv,$

where

$A = \{ (u, v) : u \gt 0, \: v \gt 0, \: u+v \lt \pi/2 \}$

has area $$\pi^2/8$$, giving $$\zeta(2) = \pi^2 / 6$$.

Categories:  #math