A limit involving the integer nearest to $n$-th power

A Pisot number will certainly work. For instance, $sqrt{2}+1$.

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.

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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 $f(t):Bbb RtoBbb R$ be a continuous function, strictly increasing for $tge a$ for some $a>0$ and unbounded for $tto+infty$ and $A$ be an open unbounded subset of $Bbb R$. Then a set $${tin (0,infty): {n: f(nt)in A } mbox{ is finite}}$$ is meager, that is a union of a countable many nowhere dense sets.

The equality $x^n=e^{nln x}$ (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 $0le s<t<ell$ a set $$X=left{x>1: frac{s}ell leleft{frac{x^n}ellright}le frac{t}ell mbox{ for each natural }nright}$$ contains a subset parameterized by an infinite binary tree with $2^{aleph_0}$ nodes, that is $|X|=2^{aleph_0}$.

Given $0le s<t<ell$, let a finite (or infinite) sequence of integers $(a_1,a_2,a_3,ldots,a_n)$ (or $(a_i)_{igeq 1}$) is good if $a_1getfrac{2ell}{t-s}$, $$left(a_iell+sright)^{frac{i+1}{i}} leq a_{i+1}ell+s< (a_{i+1}+1)ell+tle left(a_iell+tright)^{frac{i+1}{i}}label{1}tag{1}$$ for every $i$ between $1$ and $n-1$ (or every $i$ if the sequence is infinite).

Lemma 1. If $(a_i)$ is good, then $a_iell+sge left(tfrac{2ell}{t-s}right)^i$ 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 $left(a_iell+tright)^{frac{i+1}{i}}-left(a_iell+sright)^{frac{i+1}{i}}ge 2ell$ for every $i$.

Proof. $$left(a_iell+tright)^{frac{i+1}{i}}-left(a_iell+sright)^{frac{i+1}{i}}=$$ $$left(a_iell+sright)^{frac{i+1}{i}}left(left(1+frac{t-s}{ a_iell+s }right)^{frac{i+1}{i}} -1right) stackrel{mbox{(by Bernoulli's inequality)}}ge$$ $$left(a_iell+sright)^{frac{i+1}{i}}left(frac{i+1}{i}cdotfrac{t-s}{ a_iell+s }right)> left(a_iell+sright)^{frac{1}{i}}(t-s)stackrel{mbox{(by Lemma 1) }}ge 2ell.$$

Lemma 3. If a finite sequence $(a_k)_{1leq kleq i}$ is good, then there are least two good sequences of length $i+1$ that extend it.

Proof. Consider the interval $I=left[left(a_iell+sright)^{frac{i+1}{i}},left(a_iell+tright)^{frac{i+1}{i}}right]$. By the preceding lemma it has length at least $2ell$, 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 $left(a_1right)=left(tfrac{2ell}{t-s}right)$ we built an infinite binary tree, that will contain $2^{aleph_0}$ nodes.

Finally, if $a=(a_i)_{igeq 1}$ is an infinite good sequence, by construction the sequences $u_i=left(a_iell+sright)^{frac{1}{i}}$ and $v_i=left(a_iell+tright)^{frac{1}{i}}$ 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 $x_aneq x_b$ 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 $I_a=[a_iell+s, a_iell+t]$ and $I_b=[b_iell+s, b_iell+t]$ 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 $left{frac {x^n}ellright}$ is convergent, but it doesn’t work.