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?
Best Answer
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.