On the convergence of $\sum\frac{\log(n)}{n}\{x^n+x^{-n}\}$ As stated in the title, I'm trying to determine the values of $x\in \mathbb{R}$ for which
$$\sum_{n=1}^\infty \frac{\log(n)}{n}\{x^n+x^{-n}\}<+\infty$$
where $\{x\}$ is the fractional part (any convention on the fractional part of negatives numbers is accepted).
Now, $\sum_{n=1}^{\infty} \frac {\log(n)}{n}=+\infty$, so the trivial bound $\{x\}<1$ is no use.
I thought about using the Dirichlet test, but still the problem of determining whether $\sum \{x^n+x^{-n}\}$ is bounded is out of my reach.
Obviously, the problem is easily solvable if one restricts to $x \in \mathbb{Z}-\{0\}$ , but I don't see an obvious extension to the general case.
Similarly, if the series converges for $x$ it does for $\frac{1}{x}$ too.
How should I approach the problem? I feel like there's an easy solution that I am missing.
 A: Let $D=\left\{x\,|\,x>1 \text{ and } \{\operatorname{frac}(x^n)\}\text{ is equidistributed on }[0,1]\right\}$. In 1914, Hardy and Littlewood proved that $\mathbb R_{x>1}\setminus D$ has Lebesgue measure $0$.
Therefore, it can be proved that

$$S(x):=\sum_{n=1}^\infty \frac{\log(n)}{n}\operatorname{frac}(x^n+x^{-n})\text{ diverges for almost all }x\in\mathbb R$$


Proof:
For the moment only $x\in D$ is considered.
Let $c,\delta$ be real numbers satisfying $0<c<c+\delta<1$.
Let $\pi(N)$ be the counting function which counts the number of natural $n\le N$ such that $\operatorname{frac}(x^n)\in [c,c+\delta)$.
Let $m_1,m_2,\cdots$ be a sequence of natural numbers, where $m_i$ is the $i$th smallest natural number satisfying $\operatorname{frac}(x^{m_i})\in[c,c+\delta)$.
Let $S_N(x)$ be the $N$th partial sum of $S(x)$. Obviously,
$$S_N(x)>\sum^{I_N}_{i=1}\frac{\ln m_i}{m_i}\operatorname{frac}(x^{m_i}+x^{-m_i})$$ where $I_N=\max\{i\,|\, m_i\le N\}$.
We will prove this lower bound diverges.

First, we want to lower bound $\operatorname{frac}(x^{m_i}+x^{-m_i})$.
Let $x^{m_i}=Z_i+c+\Delta$ where $Z_i$ is an integer and, by definition of $m_i$, $0\le\Delta<\delta$.
Since $x^{-m_i}$ vanishes as $i\to\infty$, for sufficiently large $i\ge i_0$, we have
$$\operatorname{frac}(x^{m_i}+x^{-m_i})=x^{m_i}+x^{-m_i}-Z_i=c+\Delta+\frac1{Z_i+c+\Delta}>c$$ 
Hence, $$S_N(x)>c\sum^{I_N}_{i=i_0}\frac{\ln m_i}{m_i}$$

Next, we want to lower bound $\displaystyle{\frac{\ln m_i}{m_i}}$.
By definition of equidistribution, $$\lim_{N\to\infty}\frac{\pi(N)}{N}=\delta$$ 
Thus, by definition of limit, $$(\delta-\epsilon)N<\pi(N) \qquad (1)$$ for fixed, small positive $\epsilon<\delta$ and sufficiently large $N$.
By the definition of the sequence $m_i$, we have $\pi(m_i)=i$. Clearly $m_i$ can be arbitrarily large, hence we are allowed to set $N=m_i$ in $(1)$ to obtain
$$m_i<\frac{i}{\delta-\epsilon}$$
for sufficiently large $i$, say, $i\ge i_1$.
Therefore, $$\frac{\ln m_i}{m_i}>\frac{\delta-\epsilon}{i}\ln\frac{i}{\delta-\epsilon}>(\delta-\epsilon)\frac{\ln i}{i}$$
As a result, let $\iota=\max\{i_0,i_1\}$, then
$$S_N(x)>c(\delta-\epsilon)\sum^{I_N}_{i=\iota}\frac{\ln i}{i}\sim \frac{c}{2}(\delta-\epsilon)\ln^2(I_N)\xrightarrow{N\to\infty}+\infty$$
$S(x)$ diverges for almost all $x>1$. 

For negative $x$, since $\operatorname{frac}(\cdot)$ is always non-negative, we have
$$S_N(x)>\sum^{N/2}_{n=1}\frac{\ln 2n}{2n}\operatorname{frac}(x^{2n}+x^{-2n})>\sum^{N/2}_{n=1}\frac{\ln n}{2n}\operatorname{frac}(x^{2n}+x^{-2n})=\frac12S_{N/2}(x^2)$$ or in other words $$S_N\left(-\sqrt{|x|}\right)>\frac12 S_{N/2}(|x|)$$
Due to the bijectivity of $\sqrt{|x|}$, the conclusion extends to $x<-1$. Moreover, as $S(x)$ is invariant upon $x\to\frac1x$, it can be concluded that $S(x)$ diverges for almost all $x\in\mathbb R$. $\blacksquare$
