Let $$S:=\left\{x\in \mathbb{R}:\sum \frac{\ln(n)}{n} \{x^n+x^{-n} \}<+\infty\right\}\\ S':=S\cap(1,+\infty)$$
In this post it was proved that $\mu(S)=0$, thanks to the equidistribution of $\{x^n\}$.
Trying to understand a little bit more about the structure of $S'$ (from which the structure of $S$ follows thanks to the map $x\to \frac{1}{x}$, under which the series is invariant), I noticed the following:
Note that the related series $\sum \{x^n+x^{-n}\}^2$ converges iff $x$ is a P.V. number.
Thus, we now know that the set of P.V. numbers is contained in $S'$. The natural question, now, is whether the converse implication holds, i.e. if $\{x: x \ \text{is a P.V. number}\}=S'$.
If this were true, it would imply the longstanding problem of whether the convergence of $\sum \{x^n\}$ is enough to determine if $x$ is a p.v. number. However, since our "test for p.v.-ness" is weaker, it may be easier to find a counterexample