Answer
The sequence $a_n$ is not monotonic but bounded.
Work Step by Step
Consider $a_n=\dfrac{2^n 3^n}{n!}=\dfrac{6^n}{n!}$ for all $n \in N$
Also, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty}\dfrac{6^n}{n!}=0$
so, the sequence $a_n$ is bounded and converges to $0$
Now, $a_1= \dfrac{6^1}{1!}=6 \lt a_2=\dfrac{6^2}{2!}=\dfrac{36}{2 \cdot 1}=18$
However, $a_6= (\dfrac{6^6}{6!}) \gt a_7=(\dfrac{6^7}{7!})$
The sequence $a_n$ is not monotonic but bounded.
