Answer
Divergent
Work Step by Step
As we know that when $x \gt 0$ so, $\lim\limits_{n \to \infty} \sqrt[n] {n}=1$ and $\lim\limits_{n \to \infty} x^{1/n}=1$ and $\lim\limits_{n \to \infty} \dfrac{x^n}{n!}=0$
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{n!}{2^n 3^n}$
This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{1}{\dfrac{(2)^n(3^n)}{n!}}=\lim\limits_{n \to \infty} \dfrac{1}{\dfrac{(6)^n}{n!}}=\dfrac{1}{0}=\infty$
Hence, $\lim\limits_{n \to \infty} a_n=\infty$ and {$a_n$} is Divergent.