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Let $A\subseteq \mathbb{N}$ be a lacunary sequence (so that $A=\{a_1<a_2<\cdots\}$ and there exists some $\lambda>1$ such that $a_{n+1}/a_n\geq \lambda$ for all $n\geq 1$). Must \[\left\{ \sum_{a\in A'}\frac{1}{a} : A'\subseteq A\textrm{ finite}\right\}\] contain all rationals in some open interval?
Bleicher and Erdős conjecture the answer is no.

Steinerberger has pointed out that as written this problem is trivial: simply take some lacunary $A$ whose prime factors are restricted (e.g. $A=\{1,2,4,8,\ldots,\}$) - clearly any finite sum of the shape $\sum_{a\in A'}\frac{1}{a}$ can only form a rational with denominator divisible by one of these restricted set of primes.

This is puzzling, since Erdős and Graham were very aware of this kind of obstruction, so it's a strange thing to miss. I assume that there was some unwritten extra assumption intended (e.g. $A$ contains a multiple of every integer).

Additional thanks to: Stefan Steinerberger