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For a positive integer $n$ define $$a_n=\sum_{k=1}^n k!.$$ Prove that the set of primes that divide some $a_n$ is infinite.

Some progress I guess. Let $S$ be the set of primes dividing some $a_n$, and suppose for contradiction that $S$ is finite. There must exist a prime $p\in S$ and an $N>p$ such that $p\mid a_N$: then $p\mid a_k$ for all $k\geq N$.

Let $v_p(n)$ denote the $p$-adic valuation of $n$. If there is an $n$ such that $v_p(a_{n-1})\neq v_p(n!)$ for all primes $p\in S$, then it follows that $v_p(a_k)=\min\{v_p(a_{n-1}),v_p(n!)\}$ for all $k\geq n$, so $a_n$ is bounded, contradiction.

Else, there exist primes $p_1,\dots,p_k\in S$ such that $v_{p_i}(a_{n-1})=v_{p_i}(n!)$ for all $n$. But from here, I'm stuck.

zhw.
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jlammy
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  • What does "...divide some $a_n$..." mean? – the_candyman Apr 30 '17 at 19:38
  • i.e. $p\in S$ iff there exists an $n$ such that $p\mid a_n$ – jlammy Apr 30 '17 at 19:39
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    https://artofproblemsolving.com/community/c6h481900p2699459 Related question on MSE: https://math.stackexchange.com/questions/337053/second-part-of-the-factorial-sum-divisibility-question – Winther Apr 30 '17 at 19:56
  • it seems that I have a similar thought process as shinichiman on the last post of the AoPS thread linked by @Winther. However, I'm not sure how he derives his contradiction at the end of the second paragraph? – jlammy Apr 30 '17 at 20:06

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