Limit of factorial function: $\lim\limits_{n\to\infty}\frac{n^n}{n!}.$ I am studying for a test and I am given this problem:
$$\lim_{n\to\infty}\frac{n^n}{n!}.$$
How do I go about solving this limit? Intuitively I see how the numerator is growing much faster, but how do I express this precisely? Thanks!
 A: Hint: Note that
$$
\left.\frac{(n+1)^{n+1}}{(n+1)!}\middle/\frac{n^n}{n!}\right.=\left(1+\frac1n\right)^n
$$
Thus, the ratio of the terms tends to $e$.
A: Hint: Use Stirling's approximation:
 $$
    n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n,$$
then after simplification you'll get: $$\lim_{n\to\infty}\dfrac{e^n}{\sqrt{2\pi n}}.$$

For the intuitive part (we're just working with integers to simplify things) we have: $$\dfrac{n^n}{n!}=\dfrac{\color{#C00}n\cdot \color{green}n\cdot \color{royalblue}n\cdots\color{brown}n}{\color{#C00}1\cdot\color{green}2\cdot\color{royalblue}3\cdots \color{brown}n}\xrightarrow[n\to\infty]{}\infty.$$
A: $\ln\left(\sqrt[n]{\dfrac{n!}{n^n}}\right)$
$=\dfrac{1}{n}\ln\left(\prod\limits_{k=1}^{n}\dfrac{k}{n}\right)$
$=\dfrac{1}{n}\sum\limits_{k=1}^{n}\ln\left({0+k\dfrac{1-0}{n}}\right)$
By Riemann sum:
$\lim\limits_{n\to +\infty}\ln\left(\sqrt[n]{\dfrac{n!}{n^n}}\right) =\int\limits_0^1\ln x\,dx=\left[x\ln x-x\right]_0^1=-1-\lim\limits_{x\to 0^+}(x\ln x-x)=-1$
And so:
$\lim\limits_{n\to +\infty}\sqrt[n]{\dfrac{n!}{n^n}}=e^{-1}$
While $0<e^{-1}<1$ we have then:
$\lim\limits_{n\to +\infty}\dfrac{n!}{n^n}=0^+$
And so:
$\lim\limits_{n\to +\infty}\dfrac{n^n}{n!}=+\infty$
A: "Intuitively I see how the numerator is growing much faster, but how do I express this precisely?": use little-o notation.
Using AM-GM we get (after a slight technicality with $n$ even or odd):
$n!=(1n)(2(n-1))\ldots\leq((\frac{n+1}{2})^{2})^{\frac{n}{2}}=\frac{(n+1)^{n}}{2^{n}}$ so $n!=o((n+1)^{n})$.
Since $\lim\limits_{n\rightarrow\infty}\frac{(n+1)^{n}}{n^{n}}=e$ this show that $n^{n}=\Theta((n+1)^{n})$.
Hence $n!=o(n^{n})$.
