Basel Problem/Proof 7

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.


Proof

By Fourier Series of $x^2$, for $x \in \left[{-\pi \,.\,.\, \pi}\right]$:

$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\left({-1}\right)^n \frac 4 {n^2} \cos n x}\right)$

Setting $x = \pi$:

\(\displaystyle \pi^2\) \(=\) \(\displaystyle \frac{\pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\left({-1}\right)^n \frac 4 {n^2} \cos \pi x}\right)\)
\(\displaystyle \pi^2\) \(=\) \(\displaystyle \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\left({-1}\right)^n \left({-1}\right)^n \frac 4 {n^2} }\right)\) Cosine of Multiple of $\pi$
\(\displaystyle \pi^2\) \(=\) \(\displaystyle \frac {\pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {2 \pi^2} 3\) \(=\) \(\displaystyle 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) \(=\) \(\displaystyle \frac {\pi^2} 6\)

$\blacksquare$