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 $\sum\_{k=1}^{\infty}\frac{1}{a\_k}=+\infty$?



Is the factorization of $a_m-a_n$ affected by the fact that $\Sigma \frac{1}{a_k}<+\infty$?
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 $m$.


