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Is there a way to partition all the positive integers into an infinite number of infinite disjoint sets $S_i$ with elements $a_{ij}$, such that $\sum_{j=1}^{\infty} 1/a_{ij}$ converges to the same value for all integers i?

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up vote 3 down vote accepted

There is, here is one.

For every subset $S$ of the positive integers, let $H(S)$ denote the sum of the inverses of the elements of $S$.

First choose a subset $S_1$ which contains $1$ and such that $H(S_1)$ is a finite irrational number (for example, the set of the factorials $n!$ for $n\geqslant1$ will do). Thus, $H(S_1)=h$ for some $h\gt1$, and $S_1$ is infinite since, for every finite subset $F$, $H(F)$ is a rational number. Then construct $(S_i)_{i\geqslant2}$ recursively as follows. For every $i\geqslant1$, choose $S_{i+1}$ such that:

  1. $S_{i+1}$ and $S_1\cup\cdots\cup S_i$ are disjoint.
  2. $S_{i+1}$ contains the smallest integer $n_{i+1}$ not in $S_1\cup\cdots\cup S_i$.
  3. $H(S_{i+1})=h$.

Since $h\gt1\gt1/n_{i+1}$, conditions 2. and 3. can be met simultaneously. Since $h$ is irrational, condition 3. ensures that $S_{i+1}$ is infinite. Finally, condition 2. ensures that $S_1\cup\cdots\cup S_i$ contains every $n\leqslant i$, hence $(S_i)_{i\geqslant1}$ is indeed a partition of the positive integers.

Note: Any irrational number $h\gt1$ may be the common value of the sums $H(S_i)$.

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We claim that for any $\sigma > 1$, there exists a countable decomposition $\mathbb{N} = \bigsqcup_{j=1}^{\infty} S_j$ into infinite subsets such that $\sum_{n \in S_j} n^{-1}= \sigma$ for all $j$.

The construction is simple. First, we prove the following claim:


Lemma. Let $T \subset\mathbb{N}$ be a set such that $\sum_{n \in T} n^{-1} = \infty$. Then there exists an infinite subset $S$ of $T$ such that $\min S = \min T$ and $\sum_{n \in S} n^{-1} = \sigma$.

Proof. Clearly $T$ is an infinite set. Let $n_1$ be the smallest element in $T$. Since $\sigma > 1$, we have $0 < n_1^{-1} < \sigma$. Now suppose elements $n_1 < n_2 < \cdots < n_k$ of $T$ are chosen so that $\sum_{i=1}^{k} n_i^{-1} < \sigma$. Then we can choose the smallest integer $n = n_{k+1} > n_{k}$ in $T$ satisfying $\sum_{i=1}^{k+1} n_i^{-1} < \sigma$. Then the set $S = \{n_1, n_2, \cdots \}$ is infinite and $\min S = \min T$.

To show that $\sum_{i} n_{i}^{-1} = \sigma$, suppose not. Then $\sum_{i} n_{i}^{-1} < \sigma$ and $\sum_{n \in T-S} n^{-1} = \infty$. Then there is $n \in T - S$ such that $0 < n^{-1} < \sigma - \sum_{i} n_{i}^{-1}$. Now find $k$ such that $n_k < n < n_{k+1}$. Then $n$ is smaller than $n_{k+1}$ but we still have $n^{-1} + \sum_{i=1}^{k} n_i^{-1} < \sigma$, contradicting the minimality assumption on $n_{k+1}$. This proves the lemma.


Then we construct $S_1, S_2, \cdots$ as follows:

Let $S_1$ be a set obtained by Lemma with $T = \mathbb{N}$. Now suppose $S_1, \cdots, S_k$ are chosen. For the set $T = \mathbb{N} - (S_1 \cup \cdots \cup S_k)$, we have $\sum_{n \in T} n^{-1} = \infty - k\sigma = \infty$ and $T$ satisfies all the conditions of Lemma, and we can find $S_{j+1} \subset T$ as in Lemma.

With this construction, induction shows that $$\{ 1, \cdots, j \} \subset S_1 \cup \cdots \cup S_{j} \subset \mathbb{N}.$$ Thus we must have $\mathbb{N} = S_1 \cup S_2 \sup \cdots$, with $\sum_{n \in S_j} n^{-1} = \sigma$ as desired.

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