Basel Problem/Proof 7
Jump to navigation
Jump to search
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 $\pi$ | |||||||||||
\(\ds \pi^2\) | \(=\) | \(\ds \frac {\pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {2 \pi^2} 3\) | \(=\) | \(\ds 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) | \(=\) | \(\ds \frac {\pi^2} 6\) |
$\blacksquare$