Showing that $\frac{x!}{x^{x}}$ tends to zero as x tends to infinity The question is pretty much in the title, I'm having difficulty formally showing that $\lim\limits _{x\to\infty}\frac{x!}{x^{x}}=0$ (despite intuitively it's fairly obvious).
Thanks in advance for helping ;)
 A: We have that 
$$n! \leq \frac{n^n}{2^n} \text{ for } n \geq 6$$
so 
$$\lim_{n \to \infty} \frac{n!}{n^n} \leq \lim_{n\to \infty} \frac{1}{2^n} = 0.$$
For $x \in \mathbb{R}$, we know that $\Gamma$ function is increasing for inputs $\geq 2$, so the limit still holds (assume $x > n \geq 6$):
\begin{align}
\Gamma(n) &= (n-1)! \\
\Gamma(x) &\leq \Gamma(\lfloor x+1 \rfloor) = \lfloor x \rfloor! 
\end{align}
so
$$\lim_{x \to \infty} \frac{\Gamma(x+1)}{x^x} \leq
\lim_{x \to \infty} \frac{\lfloor x+1 \rfloor!}{\lfloor x \rfloor^{\lfloor x \rfloor}} = 
\lim_{n \to \infty} \frac{(n+1)!}{n^n} \leq 
\lim_{n\to \infty} \frac{n+1}{2^n} = 0.$$
I hope this helps ;-)
Edit: Fixed some minor issues, thanks to @julien for noticing.
A: We assume that you want to show that 
$$\lim_{n\to\infty} \frac{n!}{n^n}=0,$$
where $n$ ranges over the positive integers.
For simplicity let $n$ be even, say $n=2m$. Then
$$\frac{(2m)!}{(2m)^{2m}}=\frac{m!}{(2m)^m}\cdot \frac{(m+1)(m+2)\cdots(2m)}{(2m)^m}.$$
Note that $\dfrac{m!}{(2m)^m}\le \dfrac{m^m}{(2m)^m}=\dfrac{1}{2^m}$.
Also, $\dfrac{(m+1)(m+2)\cdots(2m)}{(2m)^m}\le 1$.
It follows that $\dfrac{(2m)!}{(2m)^{2m}}\le \dfrac{1}{2^m}$.
A small modification takes care of odd integers. There is a lot of slack.
A: So we have
$$
\frac{x!}{x^x}=\frac{\Gamma(x+1)}{e^{x\ln x}} \qquad \forall x>0.
$$
Since $\Gamma(x+1)$ and $x\ln x$ are increasing on $[1,+\infty)$, for $n_x=\lfloor x\rfloor$, we have
$$
0\leq \frac{\Gamma(x+1)}{e^{x\ln x}}\leq \frac{\Gamma(n_x+2)}{e^{n_x\ln n_x}}=\frac{(n_x+1)!}{n_x^{n_x}}.
$$
Now observe that $\lim n_x=+\infty$ as $x$ tends to $+\infty$. So it suffices to show that
$$
\lim_{n\rightarrow +\infty}\frac{(n+1)!}{n^n}=0.
$$
One way (not the best...) to do that is to consider the series
$$
\sum_{n\geq 1} a_n=\sum_{n\geq 1}\frac{(n+1)!}{n^n}.
$$
Since 
$$
\frac{a_{n+1}}{a_n}=\frac{n+2}{n+1}\left( 1-\frac{1}{n}\right)^n\longrightarrow e^{-1}<1,
$$
the ratio test tells us the series converges. So the general term tends to $0$, which is what we want.
A: Let $\displaystyle u_n=\frac{n!}{n^n}$.
By D'Alembert test 
$$\frac{u_{n+1}}{u_n}=(1+\frac{1}{n})^{-n}=e^{-n\log(1+1/n)}\sim_{+\infty}e^{-1}<1,$$
so the series $\displaystyle \sum_nu_n$ is convergent and consequently:
$$\lim\limits _{n\to\infty}\frac{n!}{n^{n}}=0.$$
A: It is easy to see that:
$$
(n!)^2 = (1 \cdot 2 \cdot \ldots \cdot n) \cdot (n \cdot (n - 1) \cdot \ldots \cdot 1)
$$
If we substitute all the $n$ factors $n (n -k)$ by their maximum value (which happens at $k^* = \frac{n}{2}$, with value $\frac{n^2}{2^2}$) we have:
$$
\begin{align*}
(n!)^2 &\le \frac{n^{2n}}{2^{2 n}} \\
n! &\le \left( \frac{n}{2} \right)^n
\end{align*}
$$
Thus:
$$
\lim_{n \rightarrow \infty} \frac{n!}{n^n}
  \le \lim_{n \rightarrow \infty} \frac{n^n}{2^n n^n} = 0
$$
Even easier ist just to use Stirling's approximation, but thet crude bound is enough.
A: Hint: What is $\log(n!)$? Then replace a sum with integrals
