proof of values of the Riemann zeta function in terms of Bernoulli numbers

This article proves part of the theorem given in the article.

Theorem 1.

For any positive integer n


where B2n is the 2nth Bernoulli numberDlmfDlmfMathworldPlanetmathPlanetmath.

Proof. The method is as follows. Using Fourier series together with inductionMathworldPlanetmath on n, we derive a formulaMathworldPlanetmathPlanetmath for the Bernoulli periodic function B2n(x) involving an infinite sum. On setting x to 0, this sum reduces to a constant times the appropriate zeta functionMathworldPlanetmath, and the result follows.

We first compute the Fourier series for B2(x). B2(x) is periodic with period 1, so


We have


so that


But then bn=cn-c-n=0 for all n, a0=0, and for n>0, an=cn+c-n=1π2n2 (where an are the coefficients of cos and bn the coefficients of sin in the Fourier series). Thus


Using this case as an inductive hypothesis, assume that for some n2


Then on (0,1)


and thus


Since n2, the sum converges absolutely, so we can move the sum outside the integralsDlmfPlanetmath, and we get

B2n(x) =(-1)n2(2n)!(2π)2(n-1)k=11k2(n-1)cos(2πkx)𝑑x𝑑x

Thus we have established this formula for all n1. Setting x=0, then, we get


or, trivially rewriting,


But clearly ζ(2n)>0 for n1, so it must be that the B2n alternate in sign, and thus


Note that as a effect of this proof, we see that the even-index Bernoulli numbers alternate in sign!

Title proof of values of the Riemann zeta functionDlmfDlmfMathworldPlanetmath in terms of Bernoulli numbers
Canonical name ProofOfValuesOfTheRiemannZetaFunctionInTermsOfBernoulliNumbers
Date of creation 2013-03-22 17:46:37
Last modified on 2013-03-22 17:46:37
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 5
Author rm50 (10146)
Entry type Proof
Classification msc 11M99