Convergence of $a_n$ given $a_{\lfloor{x^n}\rfloor}$ converges to $0$ Sequence $a_n$ has a property that for every real $x > 1$, sequence $a_{\lfloor{x^n}\rfloor}$ converges to $0$. Does that mean that $a_n$ converges to $0$? I have tried to find a counterexample, but failed at it, so I think it might be true, but I don't know how to prove it.
 A: We will use the following general observation.

Proposition. (Croft's Lemma) Let $U_1, U_2, \cdots$ be subsets of $\mathbb{R}$ such that the interior $\mathring{U}_j$ satisfies $\sup \mathring{U}_j = \infty$ for all $j \geq 1$. Then the set
  $$ \mathcal{D} = \{ r \in \mathbb{R} : \text{for each $j \geq 1$, $n r \in U_j$ holds for infinitely many $n$} \} $$
  is dense in $(0, \infty)$.

Before proving this, let us rejoice its consequence to the problem of interest.

Claim. If $a_{\lfloor x^n \rfloor} \to 0$ as $n\to\infty$ for each $x > 1$, then $a_n \to 0$.

Proof. Define $f(r) = a_{\lfloor \exp(r) \rfloor}$. Then the assumption tells that $f(nr) \to 0$ as $n\to\infty$ along $\mathbb{N}$, for each $r > 0$. Assume otherwise that $a_n \not\to 0$. Then there exists $\epsilon > 0$ such that $|a_n| > \epsilon$ holds for infinitely many $n$. Then
$$ U := \{ r : |f(r)| > \epsilon \} = \bigcup_{n \ : \ |a_n| > \epsilon} [\log n, \log(n+1)) $$
is such that its interior $\mathring{U}$ is not bounded from above. So by Prpoposition, the set
$$ \mathcal{D} = \{ r \in (0, \infty) : \text{$nr \in U$ for infinitely many $n$} \} $$
is dense in $(0, \infty)$, and in particular, non-empty. But if $r \in \mathcal{D}$, then $|f(nr)| > \epsilon$ for infinitely many $n$, and so, $f(nr) \not\to 0$, a contradiction. ////


Lemma. If $U$ is an open subset of $\mathbb{R}$ such that $\sup U = \infty$, then $\bigcup_{n\geq N} \frac{1}{n} U$ is dense in $(0, \infty)$.

Proof. Write $\mathcal{W}_N = \bigcup_{n\geq N} \frac{1}{n} U$. We aim to show that $(a, b) \cap \mathcal{W}_N$ is non-empty for each $0 < a < b$. Since $(n+1) a < n b$ for sufficiently large $n$, we can find $r > 0$ such that
$$(r, \infty) \subseteq \bigcup_{n\geq N}(n a, n b).$$
Since $U$ is not bounded above, $U \cap (r, \infty)$ is non-empty, so we can pick $x \in U \cap (r, \infty)$. Then $x \in (n a, n b)$ for some $n\geq N$, and hence $\frac{x}{n} \in (a, b) \cap \frac{1}{n} U$. ////
Proof of Proposition. Notice that
$$ \mathcal{D} = \bigcap_{j\geq 1} \bigcap_{N\geq 1}\bigcup_{n\geq N} \frac{1}{n} U_j. $$
By the lemma, $\mathcal{D}$ is a countable intersection of open dense subsets of $(0, \infty)$. Then the conclusion follows from the Baire category theorem and the lemma above. ////
A: This is only a partial answer because I will assume that 
$$
\tag{C}\forall x\gt 1, \quad \lim_{n\to +\infty}a_{\lfloor x^n\rfloor +1}=0.
$$
I do not not whether this is a consequence of the original assumption on $(a_n)_n$. 
Define a continuous function on $[1,+\infty)$ in the following way. For each integer $n\geqslant 1$, we define $f(n):=a_n$ and we interpolate $f$ linearly on $(n,n+1)$. In particular, for $x\in [n,n+1]]$, $f(x)$ belongs to the interval 
$[\min\{a_n,a_{n+1}\},\max\{a_n,a_{n+1}\}\}]$.  It thus suffices to prove that $\lim_{x\to +\infty}f(x)=0$.  
Using (C) and the construction of $f$, we know that for all $x\gt 1$, the sequence $\left(f\left(x^n\right)\right)_{n\geqslant 1}$ goes to zero as $n$ goes to infinity. Define $g(x):=f\left(e^x\right)$ for $x\geqslant 0$. Since $g(nx)\to 0$ for all positive $x$, it follows that $g(x)\to 0$ as $x$ goes to infinity hence so does $f$.
