From: Robin Chapman Newsgroups: sci.math Subject: Re: Euler products at s = 1 Date: Wed, 16 Sep 1998 08:46:23 GMT In article <6tjtep$no4$1@gannett.math.niu.edu>, rusin@vesuvius.math.niu.edu (Dave Rusin) wrote: > Robin Chapman wrote: > >Let chi be a (nontrivial) Dirichlet character and consider the infinite > >product of (1 - chi(p)/p)^{-1} over primes p. > > > >Q1. Does this product converge? > > > >Q2. Does it converge to the "right" value, i.e., to L(1, chi)? > > > >It's easy to see that the answer to Q1 is the same as that to > > > >Q3. Does the sum of chi(p)/p converge? > > Landau, Primzahlen; it's in (I love this) Band 1, Buch 2, Teil 5, > Kapitel 25, Sekt. 109. Both additive and multiplicative forms are > treated. The specific series of which you wrote is right there at the > end of the section. Thanks to Dave for the reference. For the benefit of readers I summarize Landau's argument. We need to standard lemmas, each proved by the trick of partial summation. Lemma 1 (Dirichlet's test) Let (a_n) be a sequence of complex numbers, and let A_n = a_1 + a_2 + ... + a_n. If (A_n) is a bounded sequence, and (b_n) is a decreasing sequence of real numbers with lim b_n = 0, then sum_{n=1}^infinity a_n b_n is converegent, and sum_{n=N}^infinity a_n b_n = O(b_N). Lemma 2 Let (a_n) be a sequence of complex numbers and suppose the series sum_{n=1}^infinity a_n/n^s is convergent at a real number s = s_0. Then sum_{n=1}^infinity a_n/n^s is convergent for all s >= s_0, and defines a continuous function for these s. Let Lambda denote the von Magoldt function, i.e., Lambda(n) = log p whenever n is a power of the prime p, and Lambda(n) = 0 otherwise. Then log n = sum_{rs = n} Lambda(n). Hence chi(n) log n/n = sum_{rs = n} [chi(r)Lambda(r)/r][chi(s)/s]. We sum this identity up to n = N and get sum_{n=1}^N chi(n)log n/n = sum_{r=1}^N [chi(r)Lambda(r)/r]sum_{s <= x/r}chi(s)/s. We apply Lemma 1 to the inner sum to get sum_{s=1}^M chi(s)/s = L(1, chi) + O(1/M). Using this we get sum_{n=1}^N chi(n)log n/ n = L(1,chi) sum_{r=1}^N chi(r) Lambda(r)/r + (1/x)sum_{r=1}^N chi(r) Lambda(r). The latter sum is bounded by the arithmetic function psi(N) = O(N) by Chebyshev's theorem. Also sum_{n=1}^infinity chi(n) log n/n is convergent bu Lemma 1 so we get O(1) = L(1,chi) sum_{r=1}^N chi(r)Lambda(r)/r + O(1) and as L(1,chi) is non-zero then sum_{r=1}^N chi(r)Lambda(r)/r is bounded. By Lemma 1 again S = sum_{r=2}^infinity chi(r) Lambda(r)/[r log r] is convergent. The non-zero terms of this sum are chi(p^m)/(m p^m) for p prime. This series is not absolutely converegnt, but the terms with m >= 2 do form an absolutely convergent series. We conclude that the terms with m = 1, i.e., sum_{p prime} chi(p)/p form a convergent series. Now by lemma 2 S = lim_{s -> 1+} sum_{r=1}^infinity chi(r)Lambda(r)/[r^s log r] and this series equals log L(chi,s) by the Euler product. Hence S = log L(chi, 1). We can do some re-arrangement of the sum S, as the terms for non-prime n are absolutely convergent (we take care to preserve the order of the terms with p prime). We get S = sum_{p prime} sum_{m=1}^infinity chi(p^m)/[m p^m] = sum_{p prime} -log[1 - chi(p)/p] and taking exponentials gives exp(S) = L(chi,1) = prod_p [1-chi(p)/p]^{-1} which is what we want. Robin Chapman + "They did not have proper Room 811, Laver Building - palms at home in Exeter." University of Exeter, EX4 4QE, UK + rjc@maths.exeter.ac.uk - Peter Carey, http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/rg_mkgrp.xp Create Your Own Free Member Forum