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


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 $m$.

  • 4
    $\begingroup$ So simple! Can we expect something better? For example a sequence which grows faster than a polynomial. $\endgroup$ Oct 16 '20 at 12:46
  • 2
    $\begingroup$ Good question! I'll have to think about it. Feel free to ask it as a separate question if you like. $\endgroup$
    – Tony Huynh
    Oct 17 '20 at 5:50

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