How negative can ${\sum }_{n\le x}\frac{f(n)}{n}$ be?

Abstract

Turán observed that logarithmic partial sums ${\sum }_{n\le x}\frac{f(n)}{n}$ of completely multiplicative functions (in the particular case of the Liouville function $f(n)=\lambda (n)$) tend to be positive. We develop a general approach to prove two results aiming to explain this phenomena. Firstly, we show that for every $\epsilon >0$ there exists some $x_0\ge 1,$ such that for any completely multiplicative function $f$ satisfying $-1\le f(n)\le 1$, we have $$\sum _{n\le x}\frac{f(n)}{n}\ge -\frac{1}{(\log \log x{)}^{1-\epsilon }},x\ge x_0.$$ This improves a previous bound due to Granville and Soundararajan. Secondly, we show that if $f$ is a typical (random) completely multiplicative function $f:\mathbb{N}\to \{-1,1\}$, the probability that ${\sum }_{n\le x}\frac{f(n)}{n}$ is negative for a given large $x,$ is $O(\exp (-\exp (\frac{\log x\cdot \log \log \log x}{C\log \log x}))).$ This improves on recent work of Angelo and Xu.

arXiv comment14 pages