Limit of $2^{n} n!/n^{n}$ as $n \to \infty$ Prove that the $\lim_{n \rightarrow \infty} \frac{2^{n} n!}{n^{n}} = 0$
$\rightarrow  \frac{2^{n} n!}{n^{n}} = $ $(\frac{2}{n})^{n} n!$
Its possible to say that $\lim_{n \rightarrow \infty} $$\frac{2}{n}$ is $0$ and because of this reason $\lim_{n \rightarrow \infty} $$(\frac{2}{n})^{n}$ is $0$ also ? 
And Because of this $\lim_{n \rightarrow \infty} \frac{2^{n} n!}{n^{n}} = 0$ ?
Thanks.
 A: Not sure what you can use, so try $t= e^{\log t}$ and then notice that $\log n! = \sum_{k=1}^{n} \log k \sim n \log -n +1$, compared to the integral. Then you'll be able to cancel out a few terms. Can you handle from here? 
A: Let $a_n=\dfrac{2^{n} n!}{n^{n}}$.
Then
$$
\dfrac{a_{n+1}}{a_n}
=
\dfrac{2^{n+1} (n+1)!}{(n+1)^{n+1}}
\dfrac{n^{n}}{2^{n} n!}
=
2 \left(\dfrac{n}{n+1}\right)^n
=
2 \dfrac{1}{\left(1+\dfrac{1}{n}\right)^n}
\to
\dfrac2e < 1
$$
The ratio test implies that the series $\sum a_n$ converges and so $a_n \to 0$.
You can avoid the ratio test by noting that $\dfrac2e<0.75$ and so $\dfrac{a_{n+1}}{a_n} < 0.75$ for $n \ge N$, which gives $a_n < 0.75^{n-N} a_N \to 0$.
A: Bernoulli's Inequality says that $\left(1+\frac1n\right)^n$ is an increasing sequence. Thus, for $n\ge2$,
$$
\begin{align}
\frac{a_{n+1}}{a_n}
&=\frac{\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{2^nn!}{n^n}}\\[3pt]
&=\frac2{\left(1+\frac1n\right)^n}\\
&\le\frac2{\left(\frac32\right)^2}\\
&=\frac89
\end{align}
$$
Therefore, for $n\ge2$,
$$
a_n\le a_2\left(\frac89\right)^{n-2}
$$
Thus, we can say that for $n\ge2$,
$$
0\le\frac{2^nn!}{n^n}\le2\left(\frac89\right)^{n-2}
$$
and by the Squeeze Theorem, we have
$$
\lim_{n\to\infty}\frac{2^nn!}{n^n}=0
$$
A: *

*noted: $x=2^nn!$ and $y=n^n$


1- simplification:
$\frac{ln(x)}{ln(y)}=\frac{nln(2)}{nln(n)}+\frac{ln(n!)}{ln(n^n)}$
2- property:
$\lim \frac{x}{y}=0$ means $\lim (\ln \frac{x}{y})=ln(0)=-\infty$
$lim(ln(y)-ln(x))=\infty$ $\rightarrow$ $y>x$
from the graph shown if $y>x$ then $y-ln(y)>x-ln(x)$ which means $y-x>ln(y)-ln(x)$
$lim(lny-lnx)<lim(y-x)$ $\rightarrow$ $lim(y-x)=\infty$ $\rightarrow$ $lim(lne^y-lne^x)=\infty$
$\rightarrow$ $lim (ln\frac{e^y}{e^x})=\infty$ $\rightarrow$ $lim (ln\frac{e^x}{e^y})=-\infty$ $\rightarrow$ $$lim\frac{e^x}{e^y}=0$$
3- Calculate the limit
$lim_{n->\infty} \frac{ln(2)}{ln(n)}+\frac{ln(n!)}{ln(n^n)}=0+0=0$
$lim_{n->\infty} \frac{ln(x)}{ln(y)} =0$ means
$lim_{n->\infty} \frac{e^{ln(x)}}{e^{ln(y)}} = 0$
$$lim_{n->\infty} \frac{x}{y} = 0$$
