From: Robin Chapman Subject: Re: alternating series Date: Fri, 03 Dec 1999 19:45:47 GMT Newsgroups: sci.math Keywords: values of certain Dirichlet L-functions In article <828n0n$8im$1@nnrp1.deja.com>, Serge Zlobin wrote: > Hi, > Long ago I had recieved the following result: > 1/1^3-1/2^3+1/4^3-1/5^3+1/7^3-1/8^3+...=4*sqrt(3)*(pi^3)/243. > But my proof was sophisticated and difficult. Does anybody know how to > calculate sums of that type? > Here is some information that had been gotten after the discussion in > fido7.ru.math. > Consider the sums s(k)=1/1^k-1/2^k+1/4^k-1/5^k+1/7^k-1/8^k+..., > k is odd. > Then s(1)=pi*sqrt(3)/9, s(3)=4*sqrt(3)*(pi^3)/243. > Through expiremental evalution it was established, that > s(n)=a(n)*(2*pi)^n / (3^n*sqrt(3)*n!), > where a(1)=1/2, a(3)=1, a(5)=5, a(7)=49, a(9)=809, a(11)=20317, > a(13)=722813. It seems to be true that a(n) is odd integer (when n>1) > and n divides a(n). > Searching the sequence 1,5,49,809,20317,722813 brings to the reference > about Glaisher's G numbers. This is asking for the values at odd positive integers of the Dirichlet L-function with conductor 3. Let L(s,chi) = sum_{n=1}^infinity chi(n)/n^s where chi(n) = 0, 1 or -1 according to if n = 0, 1 or 2 mod 3. Then L(k, chi) is a rational multiple of pi^k when k is odd. (This is a general property of Dirchlet L-functions, their values at positive integers k with the same parity as the conductor are rational multiples of pi^k). One can evaluate these values by adapting the usual proof of the zeta-function evaluation at even integers. Recall that pi cot(pi z) = 1/z + sum_{m=1}^infinity [1/(z + m) + 1/(z - m)]. Replace z by (z - 1/3) and expanding out each summand as a geometric series we get that the z^{2r} coefficient in the Macluarin series of pi cot(pi (z-1/3)) is a multiple of L(2r+1, chi). Presumably the Glaisher numbers occur in the Maclaurin series of this function. -- Robin Chapman http://www.maths.ex.ac.uk/~rjc/rjc.html "`Well, I'd already done a PhD in X-Files Theory at UCLA, ...'" Greg Egan, _Teranesia_ Sent via Deja.com http://www.deja.com/ Before you buy.