# Density and Sums of Reciprocals

Is there a notion of density for a (strictly increasing) sequence of natural numbers that decides whether the sum of the reciprocals of that sequence converges?

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Well, the function $A \mapsto \sum_{n \in A} \frac{1}{n}$ is a countably additive measure on $\mathbb{N}$, though I assume this isn't quite the answer you were hoping for. – Mark Jul 22 '10 at 19:19
The nicest characterization I know is the Muntz-Satz theorem (it is strange to say theorem-theorem, but that is what people call it). See wapedia.mobi/en/Müntz–Szász_theorem – Bill Johnson Jul 22 '10 at 19:26
Since the link in the above comment is not working, here is a Wikipedia link: Müntz–Szász theorem. – Martin Sleziak Jun 2 at 11:41

The notion of natural density gives a suficient condition. Namely, one can prove that if $A \subseteq \mathbb{N}$ has positive upper natural density then $\sum_{a \in A} \frac{1}{a}$ diverges. This condition is not a necessary one, though. A well-known result in number theory ascertains that the series of the reciprocals of the prime numbers diverges whereas $\pi(n) = o(n)$.
I. Erdös's brilliant proof of the divergence of $\sum_{p} \frac{1}{p}$: http://www.renyi.hu/~p_erdos/1938-13.pdf