A limit involving the integer nearest to $n$-th power Find all $x\in\mathbb{R}$ such that $$\lim_{n\to\infty}|x^n-\langle x^n\rangle|=0$$ where $\langle t\rangle$ is the integer nearest to $t$ (eg. $\langle\frac{1}{3}\rangle=0$, $\langle\frac{8}{3}\rangle=3$, $\langle k+\frac{1}{2}\rangle$ is not defined for $k\in\mathbb{Z}$).

I found this somewhere in the internet (today I searched again in IMO shortlists and didn't found, so it's probably not from there), tried to solve for a long time, but without nontrivial results (numbers $x\in\mathbb{Z}$ and $x\in(0,1)$ satisfy this, but I have no idea how to examine e.g. $x=\sqrt{2}$).
 A: Update. (2021, January 15). According to Wikipedia’s article, there are only countably many such $x$ with $|x|>1$, and among them no transcendental numbers are known.  Moreover, if the answer to a longstanding Pisot-Vijayaraghavan problem is affirmative then the set of such $x>1$ is coinsides with the set of Pisot-Vijayaraghavan numbers.

In 2011, at a conference in my institute, I met Mark Zel’dich, in whose thesis “On one analogue of the Poincaré  recurrence theorem” (in Ukrainian) is announced the following
Attraction theorem. Let $f(t):\Bbb R\to\Bbb R$ be a continuous function, strictly increasing for $t\ge a$ for some $a>0$ and unbounded for $t\to+\infty$ and $A$ be an open unbounded subset of $\Bbb R$. Then a set $$\{t\in (0,\infty): \{n: f(nt)\in A \} \mbox{ is finite}\}$$ is meager, that is a union of a countable many nowhere dense sets.
The equality $x^n=e^{n\ln x}$ (for $x>1$) and Attraction theorem applied to a function $f(t)=e^t$ and any open neighborhood $A$ of the set $\Bbb Z+1/2$ imply that a set of $x>1$ (and so also of $x<-1$), satisfying the claim, is meager.
On the other hand, following this answer by Ewan Delanoy, we can prove the next
Proposition.  For each $0\le s<t<\ell$ a set $$X=\left\{x>1: \frac{s}\ell \le\left\{\frac{x^n}\ell\right\}\le \frac{t}\ell \mbox{ for each natural }n\right\}$$ contains a subset parameterized by an infinite binary tree with $2^{\aleph_0}$ nodes, that is $|X|=2^{\aleph_0}$.
Given $0\le s<t<\ell$, let a finite (or infinite) sequence of integers $(a_1,a_2,a_3,\ldots,a_n)$ (or $(a_i)_{i\geq 1}$) is good if $a_1\ge\tfrac{2\ell}{t-s}$,  $$\left(a_i\ell+s\right)^{\frac{i+1}{i}} \leq a_{i+1}\ell+s< (a_{i+1}+1)\ell+t\le \left(a_i\ell+t\right)^{\frac{i+1}{i}}\label{1}\tag{1}$$ for every $i$ between $1$ and $n-1$ (or every $i$ if the sequence is infinite).
Lemma 1. If $(a_i)$ is good, then $a_i\ell+s\ge \left(\tfrac{2\ell}{t-s}\right)^i$ for every $i$.
Proof. Use induction on $i$ and  $a_i^{\frac{i+1}{i}} \leq a_{i+1}$.
Lemma 2. If $(a_i)$ is good then $\left(a_i\ell+t\right)^{\frac{i+1}{i}}-\left(a_i\ell+s\right)^{\frac{i+1}{i}}\ge 2\ell$ for every $i$.
Proof.
$$\left(a_i\ell+t\right)^{\frac{i+1}{i}}-\left(a_i\ell+s\right)^{\frac{i+1}{i}}=$$
$$\left(a_i\ell+s\right)^{\frac{i+1}{i}}\left(\left(1+\frac{t-s}{ a_i\ell+s }\right)^{\frac{i+1}{i}}
-1\right) \stackrel{\mbox{(by Bernoulli's inequality)}}\ge$$
$$\left(a_i\ell+s\right)^{\frac{i+1}{i}}\left(\frac{i+1}{i}\cdot\frac{t-s}{ a_i\ell+s }\right)>
\left(a_i\ell+s\right)^{\frac{1}{i}}(t-s)\stackrel{\mbox{(by Lemma 1) }}\ge  2\ell.$$
Lemma 3. If a finite sequence $(a_k)_{1\leq k\leq i}$ is good, then there
are least two good sequences of length $i+1$ that extend it.
Proof. Consider the interval $I=\left[\left(a_i\ell+s\right)^{\frac{i+1}{i}},\left(a_i\ell+t\right)^{\frac{i+1}{i}}\right]$. By the preceding lemma it has length at least $2\ell$, so there are at least two integers $a_{i+1}$ satisfying inequality \eqref{1}.
Lemma 4. There are $2^{\aleph_0}$ infinite good sequences.
Proof. Iterate the preceding lemma a countable number of times,
and starting from the good sequence $\left(a_1\right)=\left(\tfrac{2\ell}{t-s}\right)$ we built an infinite binary tree, that will contain $2^{\aleph_0}$ nodes.
Finally, if $a=(a_i)_{i\geq 1}$ is an infinite good sequence, by construction
the sequences $u_i=\left(a_i\ell+s\right)^{\frac{1}{i}}$ and $v_i=\left(a_i\ell+t\right)^{\frac{1}{i}}$ are
“adjacent” : $(u_i)$ is nondecreasing, $(v_i)$ is nonincreasing, and $u_i \leq v_i$
for every $i$. Then, we will have a real number $x_a \in X$ such that
$u_i \leq x_a \leq v_i$ for every $i$.
Now, we have $x_a\neq x_b$ if the sequences $a$ and $b$ are different. Indeed,
if $i$ is the smallest index such that $a_i \neq b_i$, then the intervals
$I_a=[a_i\ell+s, a_i\ell+t]$ and $I_b=[b_i\ell+s, b_i\ell+t]$ are disjoint, the first contains
$x_a^i$ and the other contains $x_b^i$.
Remark.  I tried to modify the construction to obtain $2^{\aleph_0}$ many $x$ such that a sequence $\left\{\frac {x^n}\ell\right\}$ is convergent, but it doesn’t work.
A: A Pisot number will certainly work. For instance, $\sqrt{2}+1$.
