The Basel Problem ζ(2)

The Basel Problem is:
$\boxed { \displaystyle \sum_{k=1}^{\infty} \frac {1}{k^2} = \frac {{\pi}^2}{6} }$

Euler was 28 years old when he proved that it converged.

The Basel Problem is also known as the Riemann Zeta function: ζ(2).

He studied the zeroes of the function sin x,
i.e. sin x= 0 for
$x=n\pi, \forall n = {0,\pm1,\pm 2, \pm 3...}$

sin x can be factorized as :
$\sin x = x.(1+\frac {x}{\pi}) .(1-\frac {x}{\pi}).(1+\frac {x}{2\pi}). (1-\frac {x}{2\pi}).(1+\frac {x}{3\pi}). (1-\frac {x}{3\pi})...$

Note 1: On the right-hand side, any factor = 0 when $x=n\pi, \forall n = 0,\pm1,\pm 2, \pm 3...$

Note 2:
$\frac {\sin x}{x} = (1+\frac {x}{\pi}) .(1-\frac {x}{\pi}).(1+\frac {x}{2\pi}). (1-\frac {x}{2\pi}).(1+\frac {x}{3\pi}). (1-\frac {x}{3\pi})...$
$\displaystyle \lim_{x \to 0} \frac {\sin x}{x} = 1$
Right-hand side also -> 1 when x -> 0.

$\frac {\sin x}{x} = (1-\frac {x^2}{1^{2}{{\pi}^2}}). (1-\frac {x^2}{2^{2}{{\pi}^2}}). (1-\frac {x^2}{3^{2}{{\pi}^2}})...$

Note: $(1+a).(1-a)= 1 - a^{2}$

From Taylor series,
$\sin x = +\frac {x}{1!} - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +...$

$\frac {\sin x}{x} = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} +...$

$(1-\frac {x^2}{1^{2}{{\pi}^2}}). (1-\frac {x^2}{2^{2}{{\pi}^2}}). (1-\frac {x^2}{3^{2}{{\pi}^2}})... = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} +...$

Product = Sum

Let’s expand the product, then compare the coefficient of the term $x^2$

Let
$a = -\frac {1}{1^{2}{{\pi}^2}}$
$b = -\frac {1}{2^{2}{{\pi}^2}}$
$c = -\frac {1}{3^{2}{{\pi}^2}}$
$d = - \frac {1}{4^{2}{{\pi}^2}}$
$(1+a.x^2)(1+b.x^2)(1+c.x^2)(1+d.x^2)...$
The coefficient for the term $(x^2)$ is (a + b + c + d +…)