Q:

Using Stirling approximation find $\lim_{n\to \infty} \frac {n!}{e^n} $

What is $\lim_{n\to \infty}\frac {n!}{e^n} ? $
e.g the expression $\frac {n!}{e^n}$ approximates to what as n gets larger?
Here I should use Stirling approximation which is $n!\approx \sqrt {2\pi n}(\frac {n}{e})^n$
How to approach now? Any suggestions?

A:

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\begin{align}
{n! \over \expo{n}} &
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\large\sim}\,\,\,
{\root{2\pi}n^{n + 1/2}\expo{-n} \over \expo{n}} =
\root{2\pi}\exp\pars{\bracks{n + {1 \over 2}}\ln\pars{n} - 2n}
\\[5mm] & =
\root{2\pi}\exp\pars{n\bracks{\ln\pars{n} - 2} + {1 \over 2}\ln\pars{n}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\large\to}\,\,\,\bbx{+\infty}
\end{align}

A:

A simpler and direct method is to note that
$$ e^e=\sum_{n=0}^\infty \frac{e^n}{n!}$$
converges, hence the summand $\frac{e^n}{n!}$ tends to $0$.

A:

A variant, using asymptotic equivalence. 
First of all, note Stirling's formula is NOT a approximation formula, in the sense that the values of $n!$ and of the formula get closer and closer. Asymptotic equivalence simply means the ratio of both tends to$1$ as $n$ tends to $\infty$
We'll find the limit of the log using equivalence and Stirling's formula:
$$\log\Bigl(\frac{n!}{\mathrm e^n}\Bigr)=\log(n!)-n\sim_\infty\log\bigl(\sqrt{2\pi}\bigr)+n\log n-2n.$$
Now, $\;\log\bigl(\sqrt{2\pi}\bigr)=o(n\log n)$ and $2n=o(n\log n)$, so
$$\log\Bigl(\frac{n!}{\mathrm e^n}\Bigr)\sim_\infty n\log n\xrightarrow[n\to\infty]{}+\infty.$$

