proof of values of the Riemann zeta function in terms of Bernoulli numbers
This article proves part of the theorem given in the article.
Proof. The method is as follows. Using Fourier series together with induction on , we derive a formula for the Bernoulli periodic function involving an infinite sum. On setting to , this sum reduces to a constant times the appropriate zeta function, and the result follows.
We first compute the Fourier series for . is periodic with period , so
But then for all , , and for , (where are the coefficients of and the coefficients of in the Fourier series). Thus
Using this case as an inductive hypothesis, assume that for some
Thus we have established this formula for all . Setting , then, we get
or, trivially rewriting,
But clearly for , so it must be that the 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 function in terms of Bernoulli numbers|
|Date of creation||2013-03-22 17:46:37|
|Last modified on||2013-03-22 17:46:37|
|Last modified by||rm50 (10146)|