From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Euler's zeta function
Date: 2 Jul 1995 03:26:30 GMT

In article <3sf207$f1f@sunforest.mantis.co.uk>,
Tony Lezard <tony@mantis.co.uk> wrote:
>In article <3sehnb$su8@highway.LeidenUniv.nl>,
>David Kleeman  <dkleeman@dds.nl> wrote:
>>Any good references on Euler's zeta function z(x)=1^(-x) + 2^(-x) + 3^(-x) + ... ?
>>I am especially interested in the proofs for z(2)=(pi^2)/6 and z(4)=(pi^4)/90.
>
>This reminds me of something I was wondering about. What's a nice short
>proof of zeta(2) = pi^2/6? All the ways of proving it that I've seen
>dive into Fourrier analysis and you need to take a lot of theory on board
>before applying it to the problem at hand. Is there are more direct
>proof?

How much cheating will you allow yourself? You could always do this:

For any polynomial P = a0 + a1 X + a2 X^2 + ... you know that the  a_i
are the elementary symmetric functions of the negatives of the roots;
thus, -a1/a0 is the sum of the reciprocals of the roots, a2/a0 is the
sum of those reciprocals taken two at a time, and so on.  From these
it is an elementary calculation to compute the sums of the powers of
the inverses of the roots; for example, their squares add up to
(a1/a0)^2 - 2 (a2/a0).

With a small leap of faith, you might decide these formulas apply
equally well to power series  P.  In particular, consider the
Taylor series at the origin for  sin(x)/x = 1 - (1/6) x^2 + ...
Clearly the function on the left has as roots the values
x = n pi  for all integers  n  other than  zero. The sum of the reciprocals
of the squares of these is of course  2 zeta(2)/pi^2; from the previous
paragraph, it's also  +1/3, so you're done.

Your ability to pass from polynomials to power series depends on
your ability to factor an analytic function as a product of linear
factors corresponding to roots; this is in fact a bit delicate, but
is covered in courses on complex analysis (see e.g. Conway's book).
Given that the method is correct, this makes the computation of
zeta(2n) pretty straightforward for each  n  by induction.

dave

==============================================================================
[Robin Chapman has a summary paper on this at his site:
	http://www.maths.ex.ac.uk/~rjc/etc/zeta2.dvi
"This gives (so far) fourteen proofs that the sum of the reciprocals of the squares of the natural numbers equals pi squared over six". ]