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Let $b$ be a fixed integer $\ge 2$ and $A$ a proper subset of $\{0, \ldots, b-1\}$. Then define $X$ to be the set of all positive integers whose base-$b$ representation consists only of digits from $A$. It's known that the series $\sum_{x \in X} \frac{1}{x}$ is convergent, see the article on Kempner series on Wiki.en and references therein, which, together with a classical result from additive-theory folklore, implies that the natural density of $X$ is zero. But what is known, say, about the upper Banach density and, for $\alpha \ge -1$, the $\alpha$-density of $X$? I seem to have a proof that these are still zero, yet would like to understand if the result is buried somewhere in the literature (but not too deep, so that someone here around can give a reference) and/or follows from relatively well-known facts and/or standard considerations.

For what it is worth, let me note that the $\alpha$-density is dominated by the natural density for $-1 \le \alpha \le 0$, so you may want to focus on $\alpha > 0$ in the above.

Just in case it may be useful, let me recall that, given $X \subseteq \mathbf N$ and a real exponent $\alpha \ge -1$, we take the natural (or asymptotic) density of $X$ to be the limit: $$\lim_{n \to \infty} \frac{\sum_{0 \ne x \in X} x^\alpha}{\sum_{x \in \mathbf N^+,\, x \le n} x^\alpha},$$ whenever this exists (this gives back the logarithmic density for $\alpha = -1$ and the natural density for $\alpha = 0$), and the upper Banach (or uniform) density of $X$ to be the limit: $$\lim_{n \to \infty} \max_{m \ge 1} \frac{|X \cap [m+1,m+n]|}{n},$$ which always exists (by Fekete's lemma on subadditive real sequences).

Note. I edited the original post to make it clear, I hope, what I mean in the comments below by saying that I'm interested in a kind of scenarios where having more or less precise information about the asymptotic behavior of the counting function of $X$ is unlikely to be useful at all (of course, I may be wrong).

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  • $\begingroup$ I should probably mention that I haven't even tried to estimate the counting function, $\pi_X$, of $X$ (which is almost surely doable), as I'm interested in a kind of scenario where the understanding of the asymptotic behavior of $\pi_X$ is not likely to help. $\endgroup$ – Salvo Tringali Jul 9 '15 at 21:41
  • $\begingroup$ All this is just tiptoeing around the well known fact that the Hausdorff (or any) dimension of X is less than 1. (Specifically, it is equal to $\log |A|/\log b$.) $\endgroup$ – Nikita Sidorov Jul 9 '15 at 22:04
  • $\begingroup$ @NikitaSidorov: Do you still think the same, after the edit? If so, I'm interested. Btw, the scenario considered in the OP is not really the most general one that I've in mind, but should be already large enough to rule out any approach purely based on the analysis of the asymptotic behavior of the counting fnc of $X$ (as per Anthony Quas' answer below). $\endgroup$ – Salvo Tringali Jul 11 '15 at 10:38
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Let $b^{k-1}<n\le b^k$. Let $m_0$ be $m$ rounded down to the nearest multiple of $b^k$. Then $|X\cap [m,\ldots,m+n-1]|\le |X\cap [m_0,\ldots,m_0+2\cdot b^k]|\le 2(b-1)^k\le 2n^\alpha$ where $\alpha=\log(b-1)/\log b$, so that $|X\cap [m,\ldots,m+n-1]|/n \le 2n^\alpha/n$. This quantity converges to 0 as $n\to\infty$.

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  • $\begingroup$ Yes, but this is precisely the kind of argument that I want to avoid (see my comment above). $\endgroup$ – Salvo Tringali Jul 9 '15 at 21:53
  • $\begingroup$ In this case, maybe it would be useful to frame the question in terms of the properties that you actually are willing to use. $\endgroup$ – Anthony Quas Jul 9 '15 at 22:07
  • $\begingroup$ I've just given it a try. $\endgroup$ – Salvo Tringali Jul 11 '15 at 6:26
  • $\begingroup$ So I'm pretty sure the same kind of arguments should work. Morally you have made a polynomial change to the weights while the set is exponentially small. My approach to this would be to replace the weights $n^{-\alpha}$ with $b^{-k\alpha}$ in the range $[b^{k-1},b^k)$. This is changing the weights by at most a bounded factor. $\endgroup$ – Anthony Quas Jul 12 '15 at 6:14

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