<s>[INST] Evaluate lim $a_n$, $a_n=\frac{3^n}{n!}$. I've come here as I need some help.

I have to  find $\lim_{x\to \infty}a_{n}$,
  where $a_{n}=\frac{3^n}{n!}$

I've tried to use Squeeze theorem ,but I couldn't find another two strings that are lower and bigger than $a_{n}$ and have the same limit.
I can consider $a_{n}\leq3^n,$ which has $\lim_{n\to \infty}3^n=\infty$,but maybe there is another proof for this exercise.
I've thought that $$\frac{3^n}{n!}=\frac{3\cdot3\cdot3\cdot3\cdots3}{1\cdot2\cdot3\cdots n}$$ but again,I do not know how to handle it.
Please give me some advice.
 [/INST] Hint. Note that for $n\geq 3$,
$$\frac{3^n}{n!}=\frac{\overbrace{3\cdot 3\cdot 3\cdot 3\cdots 3}^{n}}{1\cdot 2\cdot 3 \cdot 4\cdots n}=
\frac{3\cdot 3}{1\cdot 2}\cdot \overbrace{\frac{3}{3}\cdot\frac{3}{4} \cdots \frac{3}{n-1}}^{n-3}\cdot \frac {3}{n}\leq \frac{3\cdot 3}{1\cdot 2 }\cdot \frac {3}{n}.$$
In a similar way, for any $a>0$, for $n\geq a$,
$$\frac{a^n}{n!}\leq \frac{a^{\lfloor a\rfloor+1}}{\lfloor a\rfloor!}\cdot \frac{1}{n}.$$
</s>