Does $\sum_n x_{\lceil \alpha^n\rceil} < \infty $ imply $x_n \to 0$?

Question

Suppose we have a (nonnegative if you like) sequence $x_n$ in $\mathbb{R}$ or a general Banach space. If for all $\alpha > 1$, $\sum_{n = 1}^\infty x_{\lceil \alpha^n\rceil} < \infty$, does this imply that $x_n \to 0$?

This question is inspired by the "sub-sub-sequence" characterization of convergence (e.g. this question and duplicates). (Application: I am trying to reduce a question of stochastic sequence convergence to stochastic series convergence.)

My thoughts: the most likely counterexample to this claim would be a sequence that doesn't converge to zero, but has its fluctuations super-exponentially far apart, e.g. $x_n = 1$ if $n = m!$ for some $m$, and $0$ otherwise. In any case, if $x_n > \epsilon$ infinitely often, I am not sure whether such $n$ certainly have an infinite intersection with the indices $\{\lceil \alpha^n\rceil\}$ for sufficiently small $\alpha$. (Maybe I am missing something obvious here.)

Edit 0: The prototype sequence I have in mind is $x_n = n^{-1}$.
Edit 1: The summability criterion implies $\lim \inf x_n = 0$. Intuitively it ought to imply more, as summability is stronger than convergence to zero. Perhaps we can find a creative way to combine series at varying $\alpha$?

Answer 1

Yes, this is true! We prove two lemmas first.

Lemma 1: Suppose $1<a<b$. Then for all large enough $z \gg 0$ there is $n\in\mathbb N$ and $c\in [a, b]$ with $c^n=z$.

Proof: Let $N\in\mathbb N$ be big enough so that $a^{N+1}\leq b^N$ and note that this still holds true for any larger $N\leq n\in\mathbb N$. If $z> b^N$ then for some $n\geq N$ we have $a^n\leq z\leq a^{n+1}\leq b^n$ so that by the intermediate value theorem, there is some $c\in [a, b]$ with $c^n=z$. $\Box$

Lemma 2: For any strictly increasing sequence $(k_n)_n$ of natural numbers, there is a real $\alpha$ so that $(\lceil \alpha^n\rceil)_n$ and $(k_n)_n$ have infinitely many values in common.

Proof: Inductively, we will settle more and more binary digits of $\alpha$. At stage $n$, we have decided some $k$ many digits of $\alpha$. So we know that for some $1<a<b$ we will have $\alpha\in [a, b]$ where $a$ is the value determined so far followed by $0$'s and $b$ comes from adding infinitely many $1$'s to the representation. By Lemma 1, there is some large $k_m \gg 0$ and $c\in [a, b]$ so that $\lceil c^n\rceil=k_m$ for some $n$. Now the binary representation of $c$ agrees with the values decided so far (ignoring $1=(0.111\dots)_2$ issues) and if we add enough digits of $c$ to our list we can arrange that definitely $\lceil \alpha^n\rceil= k_m$ in the end, no matter what digits we decide on at some later stage. This completes the construction, and defines a real $\alpha$ that works. $\Box$

We are almost done. Let's prove the statement by contraposition: Suppose $(x_n)_n$ is a sequence which does not converge to $0$. So for some $\varepsilon>0$ we can find a subsequence $(x_{k_n})_n$ with $ \vert x_{k_n}\vert\geq\varepsilon$ for all $n$. By Lemma 2, there is $1<\alpha$ so that for infinitely many $m$, there is $n$ with $\lceil \alpha^n\rceil=k_m$. But then $(x_{\lceil\alpha^n\rceil})_n$ is clearly not summable!

Answer 2

I don't have enough rep, so I guess this is going to be an answer instead of a comment. I think this is probably incorrect because it seems too simple/sketchy to me, but perhaps seeing what's wrong here might be useful to people. This answer also assumes $x_n$ to be nonnegative.

We prove by contradiction. Suppose that for all $\alpha > 1,$, $\sum_{n=1}^{\infty} x_{\lceil \alpha^n \rceil} < \infty$, but $x_n \to \gamma$ for some $\gamma > 0$. In particular, this implies that for all $\epsilon > 0$ there is some $N$ so that for $n > N$, $|x_n - \gamma| < \epsilon$. Thus, there is some $N$ so that for $n > N$, we have $x_n \geq \frac{\gamma}{2}$. Fix $\alpha = 2$. Then, $$ \sum_{n=1}^{\infty}x_{2^n} = \sum_{n=1}^{\lfloor \log N \rfloor}x_{2^n} + \sum_{n=\lfloor \log N \rfloor + 1}^{\infty}x_{2^n} \geq \sum_{n=\lfloor \log N \rfloor + 1}^{\infty}x_{2^n} \geq \sum_{n=\lfloor \log N \rfloor + 1}^{\infty}\frac{\gamma}{2} $$ Since $\frac{\gamma}{2} > 0$, following the above $\sum_{n=1}^{\infty}x_{2^n} = \infty$, which is a contradiction. Thus we conclude $x_n \to 0$.