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

Abstract Turán observed that logarithmic partial sums $\sum_{n\le x}{f(n)}/{n}$ 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 there exist constants $C, x_0\ge 1,$ such that for any completely multiplicative function f satisfying $-1\le f(n)\le 1$ , we have \begin{equation*}\sum_{n\le x}\frac{f(n)}{n}\ge -\frac{C(\log\log\log{x})^2}{(\log\log{x})}, \quad x\ge x_0.\end{equation*} 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 $\sum_{n\le x}{f(n)}/{n}$ is negative for a given large x , is $O(\exp(-\exp({\log x\cdot \log\log\log x}/{C\log \log x}))).$ This improves on recent work of Angelo and Xu.