English

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

Number Theory 2022-11-11 v1 Probability

Abstract

Tur\'an observed that logarithmic partial sums nxf(n)n\sum_{n\le x}\frac{f(n)}{n} of completely multiplicative functions (in the particular case of the Liouville function f(n)=λ(n)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 ε>0\varepsilon>0 there exists some x01,x_0\ge 1, such that for any completely multiplicative function ff satisfying 1f(n)1-1\le f(n)\le 1, we have nxf(n)n1(loglogx)1ε,xx0.\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 ff is a typical (random) completely multiplicative function f:N{1,1}f:\mathbb{N}\to \{-1,1\}, the probability that nxf(n)n\sum_{n\le x}\frac{f(n)}{n} is negative for a given large x,x, is O(exp(exp(logxlogloglogxCloglogx))).O(\exp(-\exp(\frac{\log x\cdot \log\log\log x}{C\log \log x}))). This improves on recent work of Angelo and Xu.

Keywords

Cite

@article{arxiv.2211.05540,
  title  = {How negative can $\sum_{n\le x}\frac{f(n)}{n}$ be?},
  author = {Bryce Kerr and Oleksiy Klurman},
  journal= {arXiv preprint arXiv:2211.05540},
  year   = {2022}
}

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14 pages