MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|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
The nicest characterization I know is the Muntz-Satz theorem (it is strange to say theorem-theorem, but that is what people call it). Seeüntz–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 $\pi(n) = o(n)$.

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 + ...


I. Erdös's brilliant proof of the divergence of $\sum_{p} \frac{1}{p}$:

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

share|cite|improve this answer

Your Answer


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.