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?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ 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. $\endgroup$– MarkCommented Jul 22, 2010 at 19:19
-
1$\begingroup$ 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 $\endgroup$– Bill JohnsonCommented Jul 22, 2010 at 19:26
-
1$\begingroup$ Since the link in the above comment is not working, here is a Wikipedia link: Müntz–Szász theorem. $\endgroup$– Martin SleziakCommented Jun 2, 2016 at 11:41
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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.
answered Jul 22, 2010 at 18:53