Is this a series convergence criterion? (Related to the Hilbert-Waring theorem)

Question

By the set of natural numbers I will mean $\mathbb N=\{n\in\mathbb Z\,|\,n\geq0\}.$

I have come across a condition on a sequence $(a_n)_{n=0}^\infty$ of natural numbers that I feel may imply that $$\sum_{n=0}^\infty\frac1{a_n}<\infty.$$

Let's define the operation $+$ on the set $2^\mathbb N$ by $$A+B=\{a+b\,|\,a\in A,\,b\in B\}.$$ $(2^\mathbb N, +)$ is a commutative monoid with $0$, the neutral element being $\{0\},$ and the zero element being $\varnothing.$ I will use the notation $$kA:=\underbrace{A+A+\ldots+A}_k.$$

Now suppose $(a_n)_{n=0}^\infty$ is a sequence of natural numbers. Suppose $a_i=a_j\implies i=j.$ Let $A=\{a_n\}_{n=0}^\infty\cup \{0\}.$ Suppose there doesn't exist $k\geq1$ such that $kA$ is closed under addition. (Let's call this condition $(*)$.) Does it imply $$\sum_{n=0}^\infty\frac1{a_n}<\infty?$$

$0\in A$ gives $$A\subseteq 2A\subseteq 3A\subseteq\ldots$$ Clearly then, the negation of $(*)$ implies that $kA=\langle A\rangle$, that is $kA$ is the submonoid generated by $A$. That's because $A\subseteq kA\subseteq\langle A\rangle.$

I will try to explain why I think it might work. It's very weak evidence but I don't see counter-evidence.

I feel that this condition says something about the behavior of $(a_n)$ for very large $n$. I think that for the condition to hold, there must be large gaps between the elements of the sequence as $n$ gets large. If $a_n=n,$ then $1A=A=\mathbb N$ is closed under addition and $$\sum_{n=0}^\infty\frac1{a_n}=\infty.$$

More generally, by the Hilbert-Waring theorem, if $a_n=n^s$ for some natural $s\geq1,$ then there exists $k\geq1$ such that $kA=\mathbb N$. I think it means that these sequences don't grow fast enough. Of course, for $s\geq 2$, the series $\sum_{n=0}^\infty\frac1{a_n}$ converges so the converse of the statement I'm asking about is false. But I think it only means that my "conjecture" is weaker than it could be.

Under Goldbach's conjecture, if $a_n=p_{n+1},$ the $n+1$-th prime number, then $\langle A\rangle=\mathbb N\setminus\{1\}$ and already $3A=\langle A\rangle$. It is widely known that $$\sum_{n=1}^\infty\frac1{p_n}=\infty.$$

If $a_n=2^n,$ then of course $\langle A\rangle=\mathbb N$, since all natural numbers can be written in the binary system. But it's clear that no $k\geq 1$ gives $kA=\mathbb N.$ And of course $\sum_{n=0}^\infty\frac1{a_n}<\infty.$

Answer 1

Let $A$ contain enough of the first few primes $2,3,5,\dots,p_n$ so that $\sum^np_j^{-1}\gt10$. Then let it miss all the primes from $p_{n+1}$ to $2p_{n+1}$, so that $2p_{n+1}$ is not in $2A$, so $2A$ is not closed under addition. Now put in another truckload of primes, all bigger than $2p_{n+1}$, call them $p_m,p_{m+1},\dots,p_s$, with $\sum_m^sp_j^{-1}\gt10$. Then leave out all the primes from $p_{s+1}$ to $3p_{s+1}$, so $3p_{s+1}$ is not in $3A$, so $3A$ is not closed under addition. And so on. I think you can get a set of primes this way with divergent reciprocal sum, but with no $kA$ closed under addition.

EDIT: in fact, no primes are necessary. Take any sequence with divergent reciprocal sum --- for example, $1,2,3,4,5,6,7,\dots$ --- and just take enough terms to add up to 17, then leave out enough terms to create a "hole" in $2A$, then put in enough terms to add up to another 17, then leave out enough terms to create a hole in $3A$, and so on, and so on.