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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 $\sum_{n=0}^\infty\frac1{a_n}<\infty.$

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You might want to stipulate that $\lim\,a_n=\infty$, or else there could be naturals in the sequence with infinite multiplicity and it'd mess things up. Also, I'm not sure if (semigroups) and (monoid) tags are all that relevant, but (additive-combinatorics) seems like it'd be a good fit. –  anon Jul 11 '12 at 16:50
    
@anon Thanks. I was thinking about it a couple of days ago and got confused when I was typing this. I'm still not sure all the details are all right... I'll edit this one. –  user23211 Jul 11 '12 at 16:52
    
@anon I'll just require that everything occurs exactly once in the sequence, because I was thinking more about sets than sequences anyway. –  user23211 Jul 11 '12 at 17:01
    
(Error in the reasoning of my previous comment. Anyway, it was that the sequence $a_n$ that takes the value $2^r$ precisely $2^r$ times and no other value will I think have property $(*)$, but the sum of reciprocals is divergent.) –  anon Jul 11 '12 at 17:02

1 Answer 1

up vote 2 down vote accepted

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.

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