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

Published 10 Nov 2022 in math.NT and math.PR | (2211.05540v1)

Abstract: Tur\'an 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 $\varepsilon>0$ there exists some $x_0\ge 1,$ such that for any completely multiplicative function $f$ satisfying $-1\le f(n)\le 1$, we have $$\sum_{n\le x}\frac{f(n)}{n}\ge -\frac{1}{(\log\log{x}){1-\varepsilon}}, \quad x\ge x_0.$$ 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.

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