Basel Problem

#Physics

$\displaystyle \zeta(2)=\sum_{n = 0}^{\infty} \frac{1}{n^{2}}=\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots=\frac{\pi ^{2}}{6}$

• This can be proved by applying Weierstrass factorization theorem to the Taylor expansion of $\displaystyle \text{sinc}(x)$ to obtain:
  $$
  \begin{align}
  \frac{\sin x}{x}&=1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}+\ldots\\ &=\left( 1-\frac{x}{\pi} \right)\left( 1+\frac{x}{\pi} \right)\left( 1-\frac{x}{2\pi} \right)\left( 1+\frac{x}{2\pi} \right)\left( 1-\frac{x}{3\pi} \right)\left( 1+\frac{x}{3\pi} \right)\ldots \\ &= \ldots -\frac{x^{2}}{\pi ^{2}}\left(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots \right)+\ldots \\ & \rightarrow -\frac{1}{3!}=-\frac{1}{\pi ^{2}}\sum_{n = 1}^{\infty} \frac{1}{n^{2}} \\ & \rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{2}}=\frac{\pi ^{2}}{6}
  \end{align}
  $$

• Alternatively you can find an elegant solution using infinite light towers and the Inverse Pythagorean Theorem as seen in this 3b1b video

• More intuitively, you can tell that $\displaystyle \sin(x)$ is $\displaystyle x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\ldots$ and it's also equal by rules of zeroes to be $\displaystyle a(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\ldots$ where $\displaystyle a=$...