Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working through the problems in Montgomery & Vaughan's Multiplicative Number Theory. In Section 11.2 'Exceptional Zeros', Exercise 9a says that for a quadratic character $\chi$, show that for all $k\ge 0, x\ge1 $ $$ \sum_{n<x}\frac{\chi(n)}{n}(1-n/x)^k \ge \sum_{n<x}\frac{\lambda(n)}{n}(1-n/x)^k, $$ where $\lambda$ is Liouville's function. This is elementary. In part b, under the hypothesis that there exists a $k$ such that $$ \sum_{n<x}\frac{\lambda(n)}{n}(1-n/x)^k\ge0 $$ for all $x\ge 1$, (no such $k$ is known to exist), one is to show that for all quadratic $\chi$ and all $\sigma>0$ $$ L(\sigma,\chi)>0. $$ I expect one is meant to use a Mellin transform with Cesàro weighting, S 5.1 in M&V. The difficulty is that $\chi(n)/n$ are the Dirichlet series coefficients of $L(s+1,\chi)$, not $L(s,\chi)$. Thus (5.18) gives $$ L(\sigma+1,\chi)>0 $$ for all $\sigma>0$.

Am I missing something obvious? Alternately, the method of part a will show that $$ \sum_{n<x}\chi(n)(1-n/x)^k \ge \sum_{n<x}\lambda(n)(1-n/x)^k $$ so under the hypothesis that there exists a $k$ such that for all $x\ge1$ $$ \sum_{n<x}\lambda(n)(1-n/x)^k\ge0 $$ one gets $L(\sigma,\chi)>0$ for all quadratic $\chi$.

EDIT: The reason one cares which version of part a is used, is that the numerics for small $k$ and moderate $x$ indicate that positivity is at least plausible for the original part a. It is not for the revised version.

Of course, one expects that no such $k$ exists in either case. For $k=0$, the positivity of $\sum_{n<x}\lambda(n)/n$ was known as Turán's conjecture, the negativity of $\sum_{n<x}\lambda(n)$ was Pólya's conjecture. Both where disproved by Haselgrove, and the same methods should generalize for any $k$.

share|cite|improve this question
Maybe you should write to the authors. – Gerry Myerson Sep 10 '11 at 5:15
I did, no reply (yet). – stopple Sep 10 '11 at 16:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.