Basel Problem

$\zeta (2)=\sum _{n=0}^{\infty }\frac{1}{n^2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots =\frac{{\pi }^2}{6}$

• This can be proved by applying Weierstrass factorization theorem to the Taylor expansion of $\text{sinc}(x)$ to obtain:

$$\begin{array}{rl}\frac{\sin x}{x}& =1-\frac{x^2}{3!}+\frac{x^4}{5!}+\dots \\ & =\left(1-\frac{x}{\pi }\right)\left(1+\frac{x}{\pi }\right)\left(1-\frac{x}{2\pi }\right)\left(1+\frac{x}{2\pi }\right)\left(1-\frac{x}{3\pi }\right)\left(1+\frac{x}{3\pi }\right)\dots \\ & =\dots -\frac{x^2}{{\pi }^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots \right)+\dots \\ & \to -\frac{1}{3!}=-\frac{1}{{\pi }^2}\sum _{n=1}^{\infty }\frac{1}{n^2}\\ & \to \sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}\end{array}$$

• 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 $\sin (x)$ is $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$ and it’s also equal by rules of zeroes to be $a(x-\pi )(x+\pi )(x-2\pi )(x+2\pi )\dots$ where $a=$…