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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?

flag	asked Jul 22 2010 at 18:41 Nathan St. John 71●2

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 Schwarzmann Jul 22 2010 at 19:19

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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 2010 at 19:26

J. H. S. · Answer 1 · 2010-07-22 19:01:42Z UTC

The notion of natural density gives a suficient condition. Namely, one can prove that if $A \subseteq \mathbb{N}$ has positive upper n. 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 $\pi(n) = o(n)$.

In the following list you are to encounter some previous discussions on MO that are closely related to the current inquiry of yours:

[1] http://mathoverflow.net/questions/13230/erdos-conjecture-on-arithmetic-progressions

[2] http://mathoverflow.net/questions/4596/on-the-series-1-2-1-3-1-5-1-7-1-11

References

I. Erdös's brilliant proof of the divergence of $\sum_{p} \frac{1}{p}$: http://www.renyi.hu/~p_erdos/1938-13.pdf

Density and Sums of Reciprocals

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