How to prove that $sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$ without using Fourier Series

Can we prove that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ without using Fourier series?

Answer

Yes. The most common way to do this is attributed to Euler. It does still require Maclaurin series, however.

Consider the Maclaurin polynomial for $\frac{\sin x}{x}$:

$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots$$

However, note that this is a polynomial $p(x)$ with zeroes $\{\pm k\pi\;|\;k \in \Bbb N\}$, and for which $p(0) = 1$. These two properties mean that

$$\frac{\sin x}{x} = \left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{2\pi}\right)\cdots$$

And by multiplying adjacent terms,

$$\frac{\sin x}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right)\cdots$$

Equating the $x^2$ terms in the Maclaurin polynomial and its factored form yields

$$-\frac{x^2}{3!} = -x^2\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots\right)$$

And multiplying both sides by $-\frac{pi^2}{x^2}$ gives us

$$\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \cdots$$