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/ |
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