Answer
converges to $0$
Work Step by Step
As we know that when $x \gt 0$, $\lim\limits_{n \to \infty} \dfrac{x^n}{n!}=0$
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{3^n 6^n}{2^{-n} n!}$
This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{3^n 6^n}{2^{-n} n!}$
and $=\lim\limits_{n \to \infty} \dfrac{(3^n 6^n 2^n)}{n!}=\dfrac{36^n}{n!}=0$
Thus, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} converges to $0$.