Let $A=\{a\_n|n\in\mathbb{N}\}$ be a sequence of positive integers with the following properties:


1. $1<a\_1<a\_2<\cdots<a\_n<\cdots$,
2. $\frac{1}{a\_1}+\frac{1}{a\_2}+\cdots+\frac{1}{a\_n}+\cdots=\infty$.


Does there always exist a subsequence $B\subseteq A$ with $B=\{b\_n|n\in\mathbb{N}\}$ such that


1. $\frac{1}{b\_1}+\frac{1}{b\_2}+\cdots+\frac{1}{b\_n}+\cdots=\infty$,
2. All products $d\_1^{m\_1}\cdot d\_2^{m\_2}\cdots\cdot d\_k^{m\_k}$ are distinct (for all $k\in \mathbb{N}$), where $d\_1,\dots,d\_k$ are distinct elements of $B$ and $m\_1,\dots,m\_k$ are positive integers ?



Divergent sums of reciprocals with unique factorization property
No. Let $A$ be the sequence of semiprimes $pq$ with $p < q < p^2$. Since the sum of reciprocals of primes between $p$ and $p^2$ approaches $\ln \ln (p^2) - \ln \ln p = \ln 2$,


$$\sum \frac1A \sim \sum\_n \frac{1}{p\_n} \ln 2 = \infty.$$


But any $n$ elements of $B$ whose prime factors are all at most $p\_{n - 1}$ would violate the distinctness condition (because $d\_1^{m\_1}d\_2^{m\_2} \dotsm d\_n^{m\_n} = 1$ for $(m\_1, \dotsc, m\_n) \in \mathbb Z^n$ reduces to a linear system of $n - 1$ equations), so one of the first $n$ elements must be greater than $\sqrt{p\_n} \cdot p\_n$, and


$$\sum \frac1B < \sum\_n \frac{1}{\sqrt{p\_n} \cdot p\_n} < \infty.$$


