Basel Problem/Proof 7

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Theorem

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

where $\zeta$ denotes the Riemann zeta function.


Proof

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

$\ds x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$

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

\(\ds \pi^2\) \(=\) \(\ds \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos \pi x}\)
\(\ds \pi^2\) \(=\) \(\ds \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \paren {-1}^n \frac 4 {n^2} }\) Cosine of Multiple of $\pi$
\(\ds \pi^2\) \(=\) \(\ds \frac {\pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\ds \leadsto \ \ \) \(\ds \frac {2 \pi^2} 3\) \(=\) \(\ds 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) \(=\) \(\ds \frac {\pi^2} 6\)

$\blacksquare$