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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.

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    $\begingroup$ This question is more appropriate for math.stackexchange.com; mathoverflow is a site for research-level mathematics. $\endgroup$
    – user44191
    Nov 3, 2019 at 4:33
  • $\begingroup$ @user44191 This is a research of the possibilities in theory of partitions of $\mathbb N$. $\endgroup$
    – user147968
    Nov 3, 2019 at 4:37
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    $\begingroup$ For you, it's research; for us, it's an undergraduate exercise. $\endgroup$
    – Gerry Myerson
    Nov 3, 2019 at 8:52
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    $\begingroup$ @GerryMyerson I do not doubt that what you wrote is true. $\endgroup$
    – user147968
    Nov 3, 2019 at 20:38

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

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

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    $\begingroup$ Another natural example (before the question is closed): $\mathbb{N}_k=$ set of numbers with exactly $k$ prime divisors. $\endgroup$
    – Yaakov Baruch
    Nov 3, 2019 at 4:55
  • $\begingroup$ @YaakovBaruch That one seems even more "natural". $\endgroup$
    – user147968
    Nov 3, 2019 at 20:39