The series converges. We will show that $\frac{a_{n}}{n} \leq \frac{2}{3}$ for $n>4$; then the geometric series $\sum\left(\frac{2}{3}\right)^{n}$ majorizes the series. Case 1: $n$ has at least two distinct prime divisors. Then $n$ can be factored as $n=q r$ with some co-prime positive integers $q, r \geq 2$; without loss of generality we can assume $q>r$. Notice that $q \mid q$ ! and $r|r!| q$ !, so $n=q r \mid q$ !; this shows $a_{n} \leq q$ and therefore $\frac{a_{n}}{n} \leq \frac{q}{n}=\frac{1}{r} \leq \frac{1}{2}$. Case 2: $n$ is the square of a prime, $n=p^{2}$ with some prime $p \geq 3$. From $p^{2}|p \cdot 2 p|(2 p)$ ! we obtain $a_{n}=2 p$, so $\frac{a_{n}}{n}=\frac{2 p}{p^{2}}=\frac{2}{p} \leq \frac{2}{3}$. Case 3: $n$ is a prime power, $n=p^{k}$ with some prime $p$ and $k \geq 3$. Notice that $n=p^{k} \mid p \cdot p^{2} \cdots p^{k-1}$, so $a_{n} \leq p^{k-1}$ and therefore $\frac{a_{n}}{n} \leq \frac{p^{k-1}}{p^{k}}=\frac{1}{p} \leq \frac{1}{2}$.