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...}

image

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 +…)

-\frac {1}{1^{2}{{\pi}^2}}  -\frac {1}{2^{2}{{\pi}^2}}  -\frac {1}{3^{2}{{\pi}^2}} - \frac {1}{4^{2}{{\pi}^2}}  - ... = - \frac {1}{3!}

\frac {1}{1^{2}}  +\frac {1}{2^{2}}  +\frac {1}{3^{2}} +\frac {1}{4^{2}} +... = \frac { {{\pi}^2} }{6}

\displaystyle  \boxed { \sum_{k=1}^{\infty} \frac {1}{k^2} = \frac {{\pi}^2}{6} }

Ref:
The Math Girls

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