The Basel Problem

Theorem 

$\zeta (2) = \sum_{n \mathop = 1}^\infty {\dfrac 1 {n^2} } = \dfrac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.

Proof

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

$ x^2 = \dfrac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty [{({-1})^n \dfrac 4 {n^2} \cos n x}]$

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

\(\pi^2\) \(=\) \(\dfrac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty [{({-1})^n \dfrac 4 {n^2} \cos \pi x}\)]

\(\pi^2\) \(=\) \(\dfrac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty [{({-1})^n ({-1})^n \dfrac 4 {n^2} }\)] Cosine of Multiple of $\pi$

\(\pi^2\) \(=\) \(\dfrac {\pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \dfrac 1 {n^2}\)

\(\leadsto \ \ \) \(\dfrac {2 \pi^2} 3\) \(=\) \(4 \sum_{n \mathop = 1}^\infty \dfrac 1 {n^2}\)

\(\leadsto \ \ \) \(\sum_{n \mathop = 1}^\infty \dfrac 1 {n^2}\) \(=\) \(\dfrac {\pi^2} 6\)

$\blacksquare$.

From WikiProof