# Existence of infinite set of positive integers s.t sum of reciprocals is rational and set of primes dividing an element is infinite

Does there exist a sequence $$(a_i)_{i \geq 0}$$ of distinct positive integers such that $$\sum_{i\geq 0}\frac{1}{a_i} \in \mathbb{Q}$$ and $$\{ p \in \mathbb{P} \text{ }|\text{ } \exists\text{ } i\geq 0 \text{ s.t.}\text{ } p | a_i\}$$ is infinite?

Motivation:

All geometric series (corresponding to sets $$\{ 1,n,n^2,n^3,... \}$$) are rational and the terms obviously contain finitely many primes. The same is true for say, sums of reciprocals of all numbers whose prime factiorisation contains only the primes $$p_1, p_2, ...,p_k$$ : the sum is then $$\prod_{i=1}^k\left(\frac{p_i}{p_i-1}\right)$$

On the other side, series corresponding to sets $$\{1^2, 2^2, 3^2, ...\}, \{1^3,2^3,3^3,...\},\{1!,2!,3!,...\}$$ converge to $$\frac{\pi^2}{6}$$, Apery's constant and $$e$$ respectively, which are all known to be irrationals.

I am aware of the fact that if this statement is true then it has not been proven yet (since it implies that the values of the zeta function at positive integers are irrational, which to my knowledge has not been shown yet).

Any counterexamples or other possible observations (such as, instead of requiring the set of primes to be infinite, requiring that it contains all primes except a finite set)?

The set $$A=\{1\times 2, 2\times 3, ...,n\times (n+1),...\}$$

is such a set.

Note that $$\sum _1^\infty \frac{1}{n(n+1)}=1$$ and every prime number $$p$$ divides $$p(p+1)$$ which is an element of the set $$A$$.

We can take $$(a_i)$$ to be an increasing sequence of primes: Let $$a_1 = 2$$, and choose the following prime to be the smallest one for which the sum of reciprocals is strictly less than $$1$$.

The sum of reciprocals will converge to $$1$$; this is because $$\sum_p \frac1p = \infty$$:

Fact. When $$x_i > 0$$, $$x_i \to 0$$ and $$\sum x_i = \infty$$, then for every positive real number $$y$$ there exists a subsequence $$x_{i_k}$$ for which $$\sum_k x_{i_k} = y$$.

Proof. Construct the subsequence as above. Concretely, let $$S_0 = 0$$ and define recursively for $$k \geq 0$$, $$i_{k+1} = \min\{i : S_k + x_i < y \}$$ and $$S_{k+1} = S_k + x_{i_{k+1}}$$. Suppose $$\sum_k x_{i_k} = y - \epsilon$$. Then all $$x_i$$ which are less than $$\epsilon$$ appear in the subsequence. Thus $$\sum_{x_i < \epsilon} x_i \leq y - \epsilon < \infty,$$ a contradiction.

Let $$f=(f_1,f_2): \mathbb{N} \rightarrow \mathbb{N}^2$$ be a bijection.

Define $$a_n=(2^{3+f_1(n)}+1)^{f_2(n)}$$.

By Bang’s theorem (https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem), if $$n > m > 3$$, then there exists some prime $$p$$ dividing $$2^{2n}-1$$ but neither $$2^{2m}-1$$ nor $$2^n-1$$, thus $$p|2^n+1$$ but not $$p|2^m+1$$.

As a consequence, no integer can be both a power of $$2^m+1$$ and $$2^n+1$$, hence the $$a_n$$ are pairwise distinct.

Besides, $$\sum_n{a_n^{-1}}=\sum_{n \geq 1}{\sum_{m \geq 1}{(1+2^{n+3})^{-m}}}=\sum_{n \geq 1}{2^{-3-n}} \in \mathbb{Q}.$$