The Basel Problem

Theorem 

$\zeta (2)=\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}={\displaystyle \frac{{\pi }^2}{6}}$

where $\zeta$ denotes the Riemann zeta function.

Proof

By Fourier Series of $x^2$, for $x\in (-\pi ..\pi )$: 

$x^2={\displaystyle \frac{{\pi }^2}{3}}+\sum _{n=1}^{\mathrm{\infty }}[(-1{)}^n{\displaystyle \frac{4}{n^2}}\cos nx]$

Letting $x\to \pi$ from the left:

${\pi }^2$ $=$ ${\displaystyle \frac{{\pi }^2}{3}}+\sum _{n=1}^{\mathrm{\infty }}[(-1{)}^n{\displaystyle \frac{4}{n^2}}\cos \pi x$]

${\pi }^2$ $=$ ${\displaystyle \frac{{\pi }^2}{3}}+\sum _{n=1}^{\mathrm{\infty }}[(-1{)}^n(-1{)}^n{\displaystyle \frac{4}{n^2}}$] Cosine of Multiple of $\pi$

${\pi }^2$ $=$ ${\displaystyle \frac{{\pi }^2}{3}}+4\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}$

$⇝$ ${\displaystyle \frac{2{\pi }^2}{3}}$ $=$ $4\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}$

$⇝$ $\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}$ $=$ ${\displaystyle \frac{{\pi }^2}{6}}$

$◼$.

From WikiProof