Finding $\lim_{n \to \infty} \frac{\sqrt{n!}}{2^n}$ I'm looking for a way to find this limit:
$\lim_{n \to \infty} \frac{\sqrt{n!}}{2^n}$
I think I have found that it diverges, by plugging numbers into the formula and "sandwich" the result. However I can't find way to prove it.
I know that $n! \approx \sqrt{2 \pi n}(\frac{n}{e})^n $ when $n \to \infty$. (Stirling rule I think)
However, I don't know how I could possibly use it. I mean, I tried using the rule of De l'Hôpital after using that rule, but I didn't go any further.
 A: $$
\frac{\sqrt{n!}}{2^n}\sim\frac{(2\pi\,n)^{1/4}\Bigl(\dfrac{n}{e}\Bigr)^{n/2}}{2^n}=(2\pi\,n)^{1/4}\Bigl(\frac{\sqrt n}{2\sqrt e}\Bigr)^{n}.
$$
Since $\dfrac{\sqrt n}{2\sqrt e}$ and $n^{1/4}$ converge to $\infty$, so does $\dfrac{\sqrt{n!}}{2^n}$.
A: Take the infinite series $\,\displaystyle{\sum_{n=1}^\infty\frac{2^n}{\sqrt{n!}}}\,$ .
Putting $\,\displaystyle{a_n:=\frac{2^n}{\sqrt{n!}}}\,$ , we check the convergence of the above positive series by  D'Alembert's test:
$$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{\sqrt{(n+1)!}}\frac{\sqrt{n!}}{2^n}=\frac{2}{\sqrt {n+1}}\xrightarrow [n\to\infty]{} 0$$
Thus, the above series converges so
$$a_n=\frac{2^n}{\sqrt{n!}}\xrightarrow [n\to\infty]{} 0\Longrightarrow \frac{\sqrt{n!}}{2^n}\xrightarrow [n\to\infty]{}\infty$$
A: Hint: For any given $A\in\Bbb R$, find an $n$, such that
$$n!>A\cdot 4^n$$
then it will stay above, as $\frac{n!}{4^n}$ is increasing if $n>4$.
A: Note that if $a_n = \frac{\sqrt{n!}}{2^n}$ then we have 
$$ a_n^2 = \frac{n!}{4^{n}} = \frac{1\cdot2\cdot3\cdot4\cdot5\cdot\ldots\cdot n}{4\cdot4\cdot4\cdot4\cdot4\cdot\ldots\cdot4} \ge \frac{1\cdot2\cdot3\cdot4}{4\cdot4\cdot4\cdot4}\cdot1\cdot\frac{n}4 \rightarrow \infty $$ as $n \rightarrow \infty$.  Therefore $a_n \rightarrow \infty$ as well.  
