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?
 A: yes it's fine, maybe a little bit more care is needed for the $(\frac{3e}{n})^n$ term...
but you could simply say that
$\displaystyle \sum_0^{\infty} \frac{3^n}{n!}=e^3$
in particular it converges and hence the terms must go to zero.
A: 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 $n\rightarrow \infty$.
A: It's easier to show that, for $n\geq 3$, $n!\geq 3! 4^{n-3}$.
So $|a_n|<\frac{3^n}{6\cdot 4^{n-3}} = \frac{9} 2 \left(\frac{3}{4}\right)^{n-3}$
Show that $\left(\frac{3}{4}\right)^{n-3}\to 0$
