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 ?

  • 2
    $\begingroup$ Trivial remark: the converse is not true. $1/p^2$ for p prime is a sequence with a convergent sum but which has the unique factorization property. $\endgroup$
    – Asvin
    Dec 5, 2020 at 20:48
  • 2
    $\begingroup$ Is there an example when the greedy construction of $B$ (if $B\cap \{a_1,\ldots,a_{n-1}\}$ is constructed, take $a_n$ to $B$ if and only if it does not violate unique factorization) has finite sum of reciprocals? $\endgroup$ Dec 5, 2020 at 21:25
  • $\begingroup$ @FedorPetrov Good question! I couldn't construct such an example. At first sight it seems that the greedy construction gives the optimal $B$. So your question might be equivalent to the problem. $\endgroup$
    – shahram
    Dec 5, 2020 at 23:53
  • $\begingroup$ Can we show that if the greedy sum converges, then so does the original sum? $\endgroup$
    – Asvin
    Dec 6, 2020 at 0:06
  • 3
    $\begingroup$ @shahram ah, of course it is equivalent: the unique factorization condition means that the logarithms are independent over rationals, and the greedy algorithm finds the maximal weight independent subset (the basic property of a matroid). $\endgroup$ Dec 6, 2020 at 0:07

1 Answer 1


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

$$\sum \frac1B < \sum_n \frac{1}{\sqrt{p_n} \cdot p_n} < \infty.$$

  • 1
    $\begingroup$ I really enjoyed your answer! $\endgroup$
    – shahram
    Dec 6, 2020 at 7:09

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