2021/109) Find the sum of the series $\sum_{n=1}^{\infty}\frac{1}{n^2}$

We have tailor  expansion of $\sin\, x$  as  

$P(x) = \sin\,x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots  (1) $

This is a polynomial of degree infinite with zeroes at  0 and npi so

this is  $Ax(1- \frac{x}{\pi})(1+ \frac{x}{pi})(1-\frac{x}{2pi})(1-\frac{x}{2\pi})\cdots$

or $P(x) =Ax(1- \frac{x^2}{\pi^2})(1- \frac{x^2}{2^pi}^2)(1-\frac{x^2}{3^2pi^2})\cdots$

comparing above with (1) we get A = 1

So $P(x) =x(1- \frac{x^2}{\pi^2})(1- \frac{x^2}{2^pi}^2)(1-\frac{x^2}{3^2pi^2})\cdots$

The coefficient of $x^3$ is    $- \sum_{n=1}^{\infty}\frac{1}{n^2\pi^2}$

from (1)  coefficient of $x^3$ is $-\frac{1}{6}$

as both are same so  $- \sum_{n=1}^{\infty}\frac{1}{n^2\pi^2} = -\frac{1}{6}$

multiplying both sdes by $- \pi^2$ we get

 $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$