IMC 2019 · Day 2 · P7

The series (1) converges. We will show that $\frac{a_n}{n} \le \frac23$ for $n > 4$; then the geometric series $\sum \left( \frac23 \right)^n$ majorizes (1).

Case 1: $n$ has at least two distinct prime divisors. Then $n$ can be factored as $n = qr$ with some co-prime positive integers $q, r \ge 2$; without loss of generality we can assume $q > r$. Notice that $q \mid q!$ and $r \mid r! \mid q!$, so $n = qr \mid q!$; this shows $a_n \le q$ and therefore $$\frac{a_n}{n} \le \frac{q}{n} = \frac1r \le \frac12.$$ Case 2: $n$ is the square of a prime, $n = p^2$ with some prime $p \ge 3$. From $p^2 \mid p \cdot 2p \mid (2p)!$ we obtain $a_n = 2p$, so $$\frac{a_n}{n} = \frac{2p}{p^2} = \frac2p \le \frac23.$$ Case 3: $n$ is a prime power, $n = p^k$ with some prime $p$ and $k \ge 3$. Notice that $n = p^k \mid p \cdot p^2 \cdots p^{k-1}$, so $a_n \le p^{k-1}$ and therefore $$\frac{a_n}{n} \le \frac{p^{k-1}}{p^k} = \frac1p \le \frac12.$$