Jtdylktuy

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.

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 $ntoinfty$ 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 $ntoinfty$ along $mathbb{N}$, for each $r > 0$. Assume otherwise that $a_n notto 0$. 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, $f(nr) notto 0$, a contradiction. ////

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

Proof. Write $mathcal{W}_N = bigcup_{ngeq N} frac{1}{n} U$. 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

$$(r, infty) subseteq bigcup_{ngeq N}(n a, n b).$$

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 $ngeq N$, and hence $frac{x}{n} in (a, b) cap frac{1}{n} U$. ////

Proof of Proposition. Notice that
$$ mathcal{D} = bigcap_{jgeq 1} bigcap_{Ngeq 1}bigcup_{ngeq N} frac{1}{n} U_j. $$
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. ////

This is only a partial answer because I will assume that
$$
tag{C}forall xgt 1, quad lim_{nto +infty}a_{lfloor x^nrfloor +1}=0.
$$
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 $ngeqslant 1$, we define $f(n):=a_n$ and we interpolate $f$ linearly on $(n,n+1)$. In particular, for $xin [n,n+1]]$, $f(x)$ belongs to the interval
$[min{a_n,a_{n+1}},max{a_n,a_{n+1}}}]$. It thus suffices to prove that $lim_{xto +infty}f(x)=0$.

Using (C) and the construction of $f$, we know that for all $xgt 1$, the sequence $left(fleft(x^nright)right)_{ngeqslant 1}$ goes to zero as $n$ goes to infinity. Define $g(x):=fleft(e^xright)$ for $xgeqslant 0$. Since $g(nx)to 0$ for all positive $x$, it follows that $g(x)to 0$ as $x$ goes to infinity hence so does $f$.