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?
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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: Method 1
Using Stirling's approximation, you can write -$$\frac{n!}{e^n}\approx \left(\frac{n}{e^2}\right)^n\sqrt{2n\pi}\rightarrow \infty$$
Since $\frac{n}{e^2}\rightarrow \infty$ as $n \to \infty$
Method 2
$$\frac{n!}{e^n}=\underbrace{\left(\frac{1}{e}\right)\cdot\left(\frac{2}{e}\right)}_{<1}\cdot \underbrace{\left(\frac{3}{e}\right)\cdot\left(\frac{4}{e}\right)\cdot\left(\frac{5}{e}\right)\ldots \left(\frac{n-1}{e}\right)\cdot\left(\frac{n}{e}\right)}_{>1}\to \infty$$
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
$$\log\Bigl(\frac{n!}{\mathrm e^n}\Bigr)\sim_\infty n\log n\xrightarrow[n\to\infty]{}+\infty.$$
