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)?

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