Is there a lacunary sequence $A\subseteq \mathbb{N}$ (so that $A=\{a_1<a_2<\cdots \}$ and there exists some $\lambda >1$ such that $a_{n+1}/a_n\ge \lambda$ for all $n\ge 1$) such that$$\left\{\sum _{a\in A^{\prime }}\frac{1}{a}:A^{\prime }\subseteq A\text{ finite}\right\}$$contains all rationals in some open interval?

AMS 11 · solved (literature)

theorem erdos_355 :
    answer(True) ↔ ∃ A : ℕ → ℕ, IsLacunary A ∧ ∃ u v : ℝ, u < v ∧ ∀ q : ℚ, ↑q ∈ Set.Ioo u v →
      q ∈ {∑ a ∈ A', (1 / a : ℚ) | (A' : Finset ℕ) (_ : ↑A' ⊆ Set.range A)}