User:Michael Hardy/Petry on Euler's zeta function

Posted by David Petry on July 15th, 1995:

Subject: Euler's zeta function

Note that


 * $$ \sin\left(n\arcsin\left(\frac x n\right)\right) $$

is a polynomial for odd n.

[ ... ]

the power series for it starts off


 * $$ x - \frac{1 - \frac{n^2}{6}}{6} x^3 + \cdots . $$

[ ... ]

the roots of the polynomial are:


 * $$ n\sin\left(\frac{m\pi}{n}\right) \text{ for } \frac{-n}{2} < m < \frac{n}{2} $$

and we can factor the polynomial as


 * $$ x\prod_i \left( 1 - \frac{x^2}{r_i^2} \right) $$

where ri are the positive roots.

Equating the power series and the expansion of the product gives


 * $$ \frac{1 - \frac{1}{n^2}}{6} = \sum_{m=1}^{(n-1)/2} \frac{1}{\left(n\sin\left( \frac{m\pi}{n} \right)\right)^2 } $$

and note that


 * $$ \lim_{n\to\infty} n\sin\left( \frac{m\pi}{n} \right) =m\pi $$

for any fixed m.

and with just a little more work we have the desired result!