Partitions of $\mathbb N$ which generate only divergent series of reciprocals of elements of those sets

Question

This question is a result of some thinking about $\mathbb N$, divergent series and partitions of sets.

Although elementary, I am not skilled enough to answer it at the present moment.

Is it possible to partition $\mathbb N$ into an infinite number of infinite sets $N_k=\{n_{mk}: m \in \mathbb N\};k\in \mathbb N$ in such a way that all the sums $\{S_w=\sum_{v=1}^{+ \infty}\dfrac{1}{n_{vw}}:w \in \mathbb N\}$ are divergent?

Of course, repetitions of elements in sets are not allowed.

I think that an answer to this question is "No.", because I think that if $S_w=+ \infty$ for all $w \in \mathbb N$ then at least two of the sets won´t be disjoint.

This question is more appropriate for math.stackexchange.com; mathoverflow is a site for research-level mathematics. — user44191, Nov 03 '19 at 04:33
@user44191 This is a research of the possibilities in theory of partitions of $\mathbb N$. — , Nov 03 '19 at 04:37
For you, it's research; for us, it's an undergraduate exercise. — Gerry Myerson, Nov 03 '19 at 08:52

score 3 · Answer 1 · answered Nov 03 '19 at 04:40

3

If we look for the most "natural" example I propose:

Let $N_k$ consist of all numbers with 2-adic absolute value equal to $2^{-k}$.

answered Nov 03 '19 at 04:40

Bjørn Kjos-Hanssen

24,629

3

Another natural example (before the question is closed): $\mathbb{N}_k=$ set of numbers with exactly $k$ prime divisors. – Yaakov Baruch Nov 03 '19 at 04:55
@YaakovBaruch That one seems even more "natural". – Nov 03 '19 at 20:39

Partitions of $\mathbb N$ which generate only divergent series of reciprocals of elements of those sets

1 Answers1