Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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
add comment

1 Answer 1

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 $\pi(n) = o(n)$.

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

[1] Erdos Conjecture on arithmetic progressions

[2] 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

II. A. Rice, Density and substance: an investigation into the size of integer subsets.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.