TITLE: Is the factorization of $a_m-a_n$ affected by the fact that $\Sigma \frac{1}{a_k}<+\infty$?
QUESTION [5 upvotes]: I would like to ask the following.

Let $(a_n)$ be a sequence of natural numbers such that
$\sum_{k=1}^{\infty}\frac{1}{a_k}$ converges.   Is it true that for
infinitely many $m$, there is a $n<m$ such that $a_m-a_n$ has a prime
divisor greater than $m$?

In other words, is it true that if for every $m, n$, the difference $a_m-a_n$ has all its prime factors less than or equal to $m$, then $\sum_{k=1}^{\infty}\frac{1}{a_k}=+\infty$?

REPLY [10 votes]: No, this is false.  Define $a_1=1$, and for all $k \geq 2$ let $a_k = \big\lfloor \frac{k}{2}\big\rfloor^2$.  Note that $\sum_{k=1}^\infty \frac{1}{a_k}$ converges since it is equal to $1+2\sum_{k=1}^{\infty} \frac{1}{k^2}$.  On the other hand, for all $1<n<m$,
$$a_m-a_n= \Big\lfloor \frac{m}{2}\Big\rfloor^2 - \Big\lfloor \frac{n}{2}\Big\rfloor^2=\left(\Big\lfloor \frac{m}{2}\Big\rfloor+ \Big\lfloor \frac{n}{2}\Big\rfloor\right)\left(\Big\lfloor \frac{m}{2}\Big\rfloor - \Big\lfloor \frac{n}{2}\Big\rfloor\right).$$
Thus, all prime factors of $a_m-a_n$ are at most $m$.