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

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 $$n such that $$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 $$\sum_{k=1}^{\infty}\frac{1}{a_k}=+\infty$$?

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 $$1, $$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$$.