Basel Problem/Proof 9

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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

Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:

$\map f x = \begin {cases} \paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end {cases}$

From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as:

$\map f x \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n x} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n x}$


Setting $x = 0$:

\(\displaystyle \map f 0\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos 0} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin 0}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {0 - \pi}^2\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos 0} {n^2} }\) Sine of Zero is Zero
\(\displaystyle \leadsto \ \ \) \(\displaystyle \pi^2\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 2 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) Cosine of Zero is One
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 2 - \frac {\pi^2} 3\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 6\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)

$\blacksquare$


Sources