I am looking at the sequence $x_n = \{\log(n)\}$, where I am assuming (WLOG) that the base is $e$ and moreover $f:x\mapsto\{x\}$ takes $x$ to its fractional part.
I am wondering if it is true and possible to show that
\begin{equation}
A=\bigg\{n:x_n \geqslant \frac{1}{2}\bigg\}
\end{equation}
and
\begin{equation}
B=\bigg\{n:x_n < \frac{1}{2}\bigg\}
\end{equation}
are both infinite. I believe Weyl's criterion says that $x_n$ isn't equidistributed along $(0,1)$, but then it doesn't seem possible that $\{\log(n)\}$ is eventually bounded above or below by $\frac{1}{2}$. Hence I am hoping to verify my intuition.
Thanks in advance!
$\{x_n\}_{n \ge 1}$ can be unequally distributed along $(0,1)$, yet there could still be infinitely many members in $A$ and $B$. For example, the sequence of random variables $$X_n = \begin{cases} U_n, & 3 \mid n \\ U_n+1/2, & 3 \not\mid n, \end{cases}, \quad U_n \sim \operatorname{Uniform}(0,1/2)$$ is not equally distributed on $(0,1)$, but of course, $A$ and $B$ under your definitions would trivially be sets with infinitely many members.