How to prove that $\lim _ {n \to \infty} \frac{n!}{(n/3)^n} = ∞$? How to prove that
$$
\lim_{ n \to \infty} \frac{n!}{\left(\frac n 3 \right)^n} = \infty
$$
using only calculus 1 knowledge?
 A: If you don't wish to use Stirling's formula then you can exploit logarithms. Let $A=\frac{n!}{\left(\frac n 3 \right)^n}$. To show that this tends to infinity, consider instead $\log A=(\log n + \log (n-1)+\ldots )-n(\log n -\log 3)$.  Here the sum of the logs can be approximated using the integral of $\log x$ over the interval $[1,n]$ (the primitive is $x\log x -x$). Alternatively, use Stolz's criterion.
A: $$
n \gg 1
\quad\Longrightarrow\quad
{n! \over \left(n/3\right)^{n}}
\sim
{\sqrt{\vphantom{\large A}2\pi\,}\,n^{n + 1/2}{\rm e}^{-n}
 \over
 \left(n/3\right)^{n}}
=
\sqrt{\vphantom{\large A}2\pi\,}\,{n^{1/2} \over \left({\rm e}/3\right)^{n}}
$$
Since ${\rm e}/3 < 1$, $\lim_{n \to \infty}\left(e/3\right)^{n} = 0$. In
addition, $\lim_{n \to \infty}n^{1/2} = +\infty$. Then
$${\large%
\lim_{n \to \infty}{n! \over \left(n/3\right)^{n}} = +\infty}
$$
A: If $\lim_{n\to+\infty}|\frac{x_{n+1}}{x_n}|=\ell$, then $\lim_{n\to+\infty}x_n=0$ if $\ell<1$ and $\lim_{n\to+\infty}|x_n|=+\infty$ if $\ell>1$.
I your case $x_n=(\frac{3}{n})^nn!$ and $\frac{x_{n+1}}{x_n}=3(\frac{n}{n+1})^n=3\exp(n\ln(\frac{n}{n+1}))$. 
Since $n\ln(\frac{n}{n+1})\sim\frac{-n}{n+1}$, then $\lim_{n\to+\infty}\frac{x_{n+1}}{x_n}=\frac{3}{e}>1$ and $\lim_{n\to+\infty}x_n=+\infty$.
@newzad: $u_n=\frac{n!}{n^n}=\Pi_{k=1}^{n-1}\frac{n-k}{n}\not\to 1$, $\frac{u_{n+1}}{u_n}=(\frac{n}{n+1})^n=\exp(n\ln(\frac{n}{n+1}))\sim\frac{1}{e}<1$ so $\lim_{n\to+\infty}u_n=0$.
