Properties of Partitions of the Natural Numbers

Question

Let us partition the positive integers: Let $P_i$ be an a set of infinitely many unique positive integers such that the union over all $P_i$ is equal to the set containing all positive integers and each integer is in only one $P_i$. As another constraint, there are infinitely many $P_i$. Let $x_i(1)=\min(P_i)$ and $x_i(t)$ and $x_i(t+1)$ be members of set $P_i$ such that $x_i(t)<x_i(t+1)$.

Let $S_i$ be defined as the sum of the reciprocals of the integers in set $P_i$. So,

$$ S_i=\sum_{t=1}^{\infty}\frac{1}{x_i(t)} $$

My first question is: if $S_i<\infty$ for all $i$, then does the following sum always diverge?

$$ F=\sum_{i=1}^{\infty}\frac{1}{x_i(1)} $$

It seems obvious but if someone could provide a proof that would be great.

Another thought is that it seems rather difficult to find partitions such that all $S_i\rightarrow\infty$. The only one that I can think of is the simple sieve: $P_1$ being multiples of 2, $P_2$ being multiples of 3 that are not multiples of 2, $P_3$ being multiples of 5 that are not multiples of 2 and/or 3, etc. These $S_i$ can always be bounded below by a constant times the reciprocals of the all primes greater than a certain point, so they all diverge. Can anyone think of any other partitions of the integers such that all $S_i$ diverge?

Answer 1

Here is a counterexample: $$\begin{aligned} P_1 & = \begin{pmatrix} 1, & 2, & 4, & 8, & 16, & 32, & 64, & \dots \end{pmatrix} \\ P_2 & = \begin{pmatrix} 3, & 9, & 27, & 81, & 243, & 729, & 2187, & \dots \end{pmatrix} \\ P_3 & = \begin{pmatrix} 5, & 6, & 7, & 25, & 125, & 625, & 3125, & \dots \end{pmatrix} \\ P_4 & = \begin{pmatrix} 10, & 11, & 12, & 13, & 14, & 15, & 49, & \dots \end{pmatrix} \\ \vdots \end{aligned}$$ The pattern is: Each $P_i$ starts with the numbers $\leq 2^i$ that are not in any $P_j$, $j < i$, and continues as the geometric series $p_i^n$ once these numbers are exhausted, where $p_i$ is the $i$-th prime number. In this case, $x_{i+1}(1) > 2^{i}$, so the sum $F$ converges.

Answer 2

Depending on how we make our partition $P_1 \cup P_2 \cup P_3 \cup ... = \Bbb{N},$ the sum $$\sum_{i = 1}^\infty \frac{1}{x_i(1)}$$ need not diverge.

One easy example: Define $S = S_0 := {1, 3, 4, 9, 10, 12, ...}$ be the set of all positive integers that have no $2$ in their ternary expansion. For each $k \in S$, let $S_k$ be the set of all positive integers that have 2's in exactly the decimal digits that $k$ has $1$'s, and only zeroes and 1's elsewhere. Then $S_0 \cup S_1 \cup S_3 \cup S_4 \cup ...$ is a partition of $\Bbb{N}$, but $$\sum_{n = 0 \text{ or } n \in S_0} \frac{1}{x_n(1)} = \frac{1}{1} + \sum_{n \in S_0} \frac{1}{2n}$$ converges, since $\sum_{n \in S_0} \frac{1}{n}$ converges ($x_0(n) = O(n^{\log 3/\log 2})$ can be bounded by a $p$-series).

Answer 3

Because the harmonic series diverges, you can start at any point in the positive integers and take a sufficiently long subsequence with reciprocals summing above any value. So assign integers to the $P_i$ in stages as follows: for each $n=1,2,3,\ldots$, append enough terms to $P_1,P_2,\ldots,P_n$ so that each corresponding $S$ sums to at least $n$. Each $P_i$ is lengthened at every stage $n \ge i$, and each sum $S_i$ grows without bound; so each $P_i$ is infinite and all the $S_i$ diverge. Note that you can start with any fixed permutation of the integers and carry out this same process. So this generates uncountably many distinct partitions meeting the requirement.