Answer
Converges to $0$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{3^n 6^n}{2^{-n} n!}$
Since, $\lim\limits_{n \to \infty} \dfrac{x^n}{n!}=0$
So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{3^n 6^n}{2^{-n} n!}=\lim\limits_{n \to \infty} \dfrac{(3 \cdot 2 \cdot 3 \cdot 2)^n}{n!}=\dfrac{36^n}{n!}$
or, $=0$
Hence, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent and converges to $0$
