<s>[INST] Convergence of Sequence with factorial I want to show that
$$
 a_n = \frac{3^n}{n!}
$$
converges to zero. I tried Stirlings formulae, by it the fraction becomes
$$
 \frac{3^n}{\sqrt{2\pi n} (n^n/e^n)}
$$
which equals
$$
 \frac{1}{\sqrt{2\pi n}} \left( \frac{3e}{n} \right)^n
$$
from this can I conclude that it goes to zero because $\frac{3e}{n}$ and $\frac{1}{\sqrt{2\pi n}}$ approaching zero?
 [/INST] Alternatively to Stirling:
We have
\begin{align}
a_n =\frac{3^n}{n!}
\end{align}
Now let $n>3$, then
\begin{align}
0\leq a_n &=\frac{3^n}{n!} = \frac{3\cdot3 \cdot 3}{1\cdot 2\cdot 3}\cdot \frac{3^{n-3}}{4\cdot 5 \cdot ...\cdot n}=\frac{3\cdot3 \cdot 3}{1\cdot 2\cdot 3}\cdot \frac{3\cdot 3 \cdot ... \cdot 3}{4\cdot 5 \cdot ...\cdot n}\\
&\leq \frac{9}{2}\cdot \frac{3\cdot 3 \cdot ... \cdot 3}{4\cdot 4 \cdot ...\cdot 4} = \frac{9}{2} \cdot \Bigr(\frac{3}{4}\Bigl)^n\rightarrow0 \text{ as } n\rightarrow \infty
\end{align}
So we have $a_n\rightarrow 0$ as $n\rightarrow \infty$.
</s>