Finding $\lim_{n\to\infty}\frac{2^n\cdot n!}{n^n}$ 
$\lim_{n\to\infty}\frac{2^n\cdot n!}{n^n}$

I think we can do it by squeeze rule but I could not find a sequence bigger than this one converging to 0 without using Stirling approximation of n!
 A: Being a sequence of positive terms you can use D'Alambert's criterion.
$$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}\frac{n^n}{2^n n!}=2\left(\frac{n}{n+1}\right)^n\longrightarrow 2e^{-1}<1$$
and so $a_n=\frac{2^n n!}{n^n}\longrightarrow0$.
A: $$
\begin{align}
\lim_{n\to\infty}\frac1n\log\left(\frac{2^nn!}{n^n}\right)
&=\log(2)+\lim_{n\to\infty}\frac1n\sum_{k=1}^n\log\left(\frac kn\right)\\
&=\log(2)+\int_0^1\log(x)\,\mathrm{d}x\\[9pt]
&=\log(2)-1
\end{align}
$$
Therefore,
$$
\lim_{n\to\infty}\left(\frac{2^nn!}{n^n}\right)^{1/n}=\frac2e
$$
and so
$$
\lim_{n\to\infty}\left(\frac{2^nn!}{n^n}\right)=0
$$
A: $$\dfrac{1+2+\cdots+n}{n}\geq\sqrt[n]{n!}$$
then
$$\sqrt[n]{\dfrac{2^nn!}{n^n}}=\dfrac{2}{n}\sqrt[n]{n!}<\dfrac{n+1}{n}\to1$$
as $n\to\infty$. This shows the series $\displaystyle\sum{\dfrac{2^nn!}{n^n}}$ is converge and then $\displaystyle\lim_{n\to\infty}{\dfrac{2^nn!}{n^n}}=0$.
A: We observe:
$\sum_{n=1}^\infty \frac{2^{n}\cdot n!}{n^n}$ 
Since, $\frac{2^{n+1}(n+1)!\cdot n^n}{(n+1)^{n+1}\cdot 2^n\cdot n!}=2\cdot (\frac{n}{n+1})^n=2(1-\frac{1}{n+1})^n\to 2e^{-1}<1$ (Using the ratio test) we conclude that the series converges. Therefore:
$\lim_{n\to\infty} \frac{2^n\cdot n!}{n^n}=0$
A: Hint:  use Stirling's approximation for $n!$
A: I have a nontraditional answer that avoids any knowledge of $(1+1/n)^n$ (at least directly).
Note that for each $i = 1, \ldots, n$ there is a unique integer $1\le j \le 6$ such that $(j-1)n/6 < i \le jn/6$.  For any given $j$ there are at most $k = \lceil n/6 \rceil$ corresponding $i$.  If we bound  each $i$ from above by the corresponding multiple of $n/6$, this gives us:
$$
n! \le \left(\frac{n}{6}\right)^n2^k3^k4^k5^k6^k = \left(\frac{n}{6}\right)^n (6!)^k
$$
Now we have 
$$\frac{2^nn!}{n^n} \le \frac{2^n}{n^n} \left(\frac{n}{6}\right)^n (6!)^k
= \frac{(6!)^k}{3^n} = 3^{6k-n}\left(\frac{(6!)}{3^6}\right)^k \le 3^5\left(\frac{720}{729}\right)^k\,.$$
Here we use that $6k-n = 6\lceil n/6\rceil - n \le 5$.
Note that this converges to 0 as $n\to \infty$, as $n\to \infty$ implies $k \to \infty$.
