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?
 A: 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
$$\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \cdots$$
A: Yes. You are hereby awarded this Wikipedia article: http://en.wikipedia.org/wiki/Basel_problem.
A: This question Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ already contains proofs that does not rely   on Fourier series. 
Also 
Robin Chapman has a collection of proofs on his homepage.
