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Let $S \subset \mathbb N$ and $f(S) = \sum_{x \in S} \frac{1}{x}$. Is there some notion of the 'size' of $S$ that lets us determine whether $f$ converges or diverges?

Clearly, if $S$ is finite, $f$ converges. If $S = \mathbb N$, it diverges. But for instance, the if $S$ is the set of all prime numbers, then the sum diverges, but if it is the sum of all square numbers, it converges. This can be thought of as "there are 'less' square numbers than there are prime numbers." However, both those sets are countably infinite.

So how do we 'measure' a 'size' for these subsets? Is there some 'threshold size' where the sets go from converging to diverging?

Connor W
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The Infinite One
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    It could be fun to consider $\inf {p \mid \sum_{n \in S} n^{-p} \lt \infty}$. – JonathanZ Apr 23 '23 at 19:00
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    For the ones that diverge, you could study the rate of divergence (i.e. the asymptotic growth rate of $g(n)=\sum_{x\in S,x<n}\frac1x$). – Karl Apr 23 '23 at 19:32
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    It has nothing to do with the "size" of the infinite set $S\subset \mathbb N$ since in all cases $S$ is countably infinite, rather it has to do with the growth rate of $x_n$ (and correspondingly the decay rate of $\frac{1}{x_n}$) where $S={x_1, x_2, x_3,...}$ and the related sum is $\sum\limits_{n=1}^\infty \frac{1}{x_n}$. The Riemann zeta function $\zeta(s)=\sum\limits_{n=1}^\infty \frac{1}{n^s}$ converges for $\Re(s)>1$, which explains why it converges for $s=2$. – Steven Clark Apr 23 '23 at 20:23
  • In my comment above I forgot to state the assumption the set $S$ is ordered such that $x_n<x_{n+1}$, – Steven Clark Apr 23 '23 at 21:10
  • @StevenClark - Those are all good comments, but because the question limits itself to positive numbers, the order of elements doesn't affect convergence, and because they are whole numbers, it "feels" like that in order for the sum to converge, they have to be sparse in some sense, and that larger sets (judged by inclusion) are less likely to converge. I think it's a question that comes from an unusual angle, and I'm curious to see what people will propose. – JonathanZ Apr 23 '23 at 23:39
  • @JonathanZsupportsMonicaC I clarified the infinite set $S\subset\mathbb{N}$ where $S={x_1, x_2, x_3, ...}$ is ordered in the context of my comment on the growth rate of $x_n$ (and corresponding decay rate of $\frac{1}{x_n}$). I believe such an ordered infinite set $S$ can be shown to converge if for some $n\in\mathbb{N}$ and some $m\in\mathbb{N}$ and all integers $k\ge 0$ it can be shown that $\frac{1}{x_{n+k}}<\frac{1}{(m+k)^{1+\epsilon}}$ where $\epsilon>0$. – Steven Clark Apr 24 '23 at 01:56
  • You might be interested in the concept of "natural density" as described at https://en.wikipedia.org/wiki/Natural_density – Gerry Myerson May 08 '23 at 02:47
  • Note that the question of divergence/convergence of $\sum(1/n_j)$ depends only on the "tail" of the sequence. That is, you can change the first million terms of the sequence, or the first billion, or trillion, to whatever you like without affecting the convergence/divergence. – Gerry Myerson May 08 '23 at 02:54

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