Summations of n^(-2k)

Date: 09/10/2000 at 00:37:43
From: Mark Rosel
Subject: Summations: What are sigma(n^-2) and sigma(n^-4)

I am trying to find out what the following are:

     sigma(n^-2)
     sigma(n^-4)
     sigma(n^-[2k])
     sigma(n^-[2k+1])

where k is an integer and all are summations from 1 to infinity.

Thank you.

Date: 09/15/2000 at 13:36:33
From: Doctor Schwa
Subject: Re: Summations: What are sigma(n^-2) and sigma(n^-4)

A good discussion can be found in George Polya's excellent book, 
_Mathematics and Plausible Reasoning_.

These first two were solved by Euler. Explanation of one method for 
the first series can be found in "Infinite Series Involving Pi" at:

   http://mathforum.org/dr.math/problems/tak12.16.97.html   

and it generalizes well to any even power.

Another method, one that I prefer, is:

Consider the function sin (pi*x). (I know you're asking, "what does 
sin(pi x) have to do with the sum of 1/n^2?" but bear with me.)

On the one hand, it's a function that's zero for every integer k. So, 
as an infinite polynomial product, it must be true that

     sin(pi x) = a x (1-x)(1+x)(1-x/2)(1+x/2)(1-x/3)(1+x/3) ...

so that the right-hand side has zeros in all the right places. By 
difference of squares, we get:

     sin(pi x) = a x (1-x^2)(1-x^2/4)(1-x^2/9) ...

Now the left side can also be expanded in a power series:

     sin(pi x) = (pi x) - (pi x)^3 / 3! + ...

Using that, we first discover that a = pi to make the x coefficients 
match up, and then discover that the sum of 1/n^2 is pi^2/6 to make 
the x^3 coefficients match up.

Feel free to write back with some questions about all the steps I 
skipped in this derivation. It's a really neat problem.

And yet one more method, posted here by Dr. Rob, can be found in "Sum 
of 1/n^2" at:

   http://mathforum.org/dr.math/problems/schmidt.7.24.00.html   

The odd powers are harder. In fact as far as I know they're all still 
unsolved.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/