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Euler's proof of the Basel problem

The Basel Problem asks for the precise sum of the following infinite series:

\(\displaystyle\sum_{n=1}^{\infty}{\frac{1}{n^2}} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots\)

 

Euler's proof :

Consider the following Taylor series expansion of the sine function

\({\mbox{sin}}\ x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots\)

Dividing both sides by \(x\) we have

\(\frac{{\mbox{sin }x}}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots\)

Note that the coefficient of \(\frac{{\mbox{sin }x}}{x}\) in the above equation is \(-\frac{1}{6}\)

 

Also, note that \(\frac{{\mbox{sin }x}}{x}\) has zeroes at \(x = \pm k\pi\) for \(k = 1, 2, 3, \dots\)

Hence, by using Weierstrass factorization theorem, we have this equivalent representation of \(\frac{{\mbox{sin }x}}{x}\)

\(\frac{{\mbox{sin }x}}{x} = (1 - \frac{x}{\pi})(1 + \frac{x}{\pi})(1 - \frac{x}{2\pi})(1 + \frac{x}{2\pi})(1 - \frac{x}{3\pi})(1 + \frac{x}{3\pi}) \dots\)

\(\frac{{\mbox{sin }x}}{x} = (1 - \frac{x^2}{\pi^2})(1 - \frac{x^2}{4\pi^2})(1 - \frac{x^2}{9\pi^2}) \dots\)

Note that the coefficient of \(x^2\) in the this represenation is 

\(-( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \dots)\)

 

The coefficients of \(x^2\)in the above two represenations must be equal. So we have the following equation

\(-\frac{1}{6} = -( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \dots)\)

Hence

\(\displaystyle\sum_{n=1}^{\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}\)

 

Euler's proof of the Basel problem is one of the most beautiful and elegant proofs in the entire history of mathematics. Read this wikipedia article for historical background.

To make sure you understood the above proof, try applying it to evaluate Riemann zeta function at all positive even integers.

 

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