Show that $\frac{3^n}{n!}$ converges to $0$

Hint:
You can use the squeezing principle:

Set $u_n=\dfrac{3^n}{n!}$. Show that $\;\dfrac{u_{n+1}}{u_n}\le\dfrac34$ for all $n\ge 3$. Deduce that $$u_n\le \Bigl(\frac34\Bigr)^{n-3}u_3\enspace\text{ for all }\;n\ge 3.$$