Basel problem

Prove $$\zeta(2)=\sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}.$$ There are many proofs on this. Here we present 3 proofs.

Euler's method
Having learnt factorization, we may ask: can we "factorize" the sine function? It turns out we can. Note that the roots to the equation $\large \frac{\sin x}{x}=0$ are $n\pi$, where $n=\pm 1,\pm 2, \pm 3, ...$ We can then express $\large \frac{\sin x}{x}$ as an infinite product of linear factors $\large (1+\frac{x}{\pi})(1-\frac{x}{\pi})(1+\frac{x}{2\pi})(1-\frac{x}{2\pi})(1+\frac{x}{3\pi})(1-\frac{x}{3\pi})...,$ or equivalently $\large (1-\frac{x^2}{\pi^2})(1-\frac{x^2}{2^2\pi^2})(1-\frac{x^2}{3^2\pi^2})...(1-\frac{x^2}{n^2\pi^2})...$
From this expression, the coefficient of $x^2$ is $\large -\frac{1}{\pi^2}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}+...)=-\frac{1}{\pi^2}\sum\limits_{n=1}^\infty \frac{1}{n^2}$
Since $\large \frac{\sin x}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-...$, we can then conclude that $\large \frac{-1}{\pi^2}\sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{-1}{3!}\Rightarrow \sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$.

Fourier Series
Prerequisites:
$\large{f(x)=\frac{a_0}{2}+\sum\limits_{n=1}^\infty(a_n \cos nx+b_n \sin nx)\:\: -\pi \leq x \leq \pi\\
a_0=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\:dx\\
a_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos nx\:dx\\
b_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin nx\:dx}$
[Explanations: later]

Parserval's Identity [Proof: later]
$\large \frac{1}{\pi}\int_{-\pi}^\pi |f(x)|^2 dx=\frac{a_0^2}{2}+\sum\limits_{n=1}^\infty (a_n^2+b_n^2)$

---

Consider the Fourier series of $f(x)=x,\:\: 0 \leq x \leq 2\pi$.
$\begin{align}a_0&=\frac{1}{\pi}\int_0^{2\pi} x\:dx=\frac{1}{\pi}\frac{x^2}{2}|_0^{2\pi}=2\pi\\
a_n&=\frac{1}{\pi}\int_0^{2\pi} x \cos nx\:dx\\
&=\frac{1}{\pi}\int_0^{2\pi} x\:d(\frac{\sin nx}{n})\\
&=\frac{1}{\pi}(x\frac{\sin nx}{n}|_0^{2\pi}-\int_0^{2\pi} \frac{\sin nx}{n}\:dx)\\
&=\frac{1}{\pi}\frac{\cos nx}{n^2}|_0^{2\pi}\\
&=0\\
b_n&=\frac{1}{\pi}\int_0^{2\pi} x \sin nx\:dx\\
&=\frac{1}{\pi}\int_0^{2\pi} x\:d(-\frac{\cos nx}{n})\\
&=\frac{1}{\pi}(-x\frac{\cos nx}{n}|_0^{2\pi}+\int_0^{2\pi} \frac{\cos nx}{n}\:dx)\\
&=\frac{1}{\pi}(\frac{-2\pi\cos 2n\pi}{n}+\frac{\sin nx}{n^2}|_0^{2\pi})\\
&=-\frac{2}{n}
\end{align}$

Thus, the Fourier series of x is:
$\large \frac{2\pi}{2}+\sum\limits_{n=1}^\infty(\frac{-2}{n})\sin nx$

Using Parserval's Identity,
$\large \frac{1}{\pi}\int_0^{2\pi}x^2dx=\frac{(2\pi)^2}{2}+\sum\limits_{n=1}^\infty(\frac{-2}{n})^2$.
Since $\large \frac{1}{\pi}\int_0^{2\pi}x^2dx=\frac{1}{\pi}\frac{x^3}{3}|_0^{2\pi}=\frac{8\pi^2}{3}$,
we have $\large \frac{8\pi^2}{3}=2\pi^2+4\sum\limits_{n=1}^\infty \frac{1}{n^2} \Rightarrow \sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$

Sidenote:
Consider the Fourier series of $x^2$ and $x^k$.
We have $\large \sum\limits_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$ and $\large \zeta(2k)=\sum\limits_{n=1}^\infty \frac{1}{n^{2k}}$ respectively.

Probabilistic approach
[later]

Key points:
Sine function can be expressed as an infinite product of linear factors.
Riemann zeta function