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

Question

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?

Answer 1

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 $$\log\Bigl(\frac{n!}{\mathrm e^n}\Bigr)\sim_\infty n\log n\xrightarrow[n\to\infty]{}+\infty.$$

Answer 2

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

$$\frac{n!}{e^n}=\underbrace{\left(\frac{1}{e}\right)\cdot\left(\frac{2}{e}\right)}_{<1}\cdot \underbrace{\left(\frac{3}{e}\right)\cdot\left(\frac{4}{e}\right)\cdot\left(\frac{5}{e}\right)\ldots \left(\frac{n-1}{e}\right)\cdot\left(\frac{n}{e}\right)}_{>1}\to \infty$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 4

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 $0$.