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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.

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up vote 8 down vote accepted

$$ \frac{\sqrt{n!}}{2^n}\sim\frac{(2\pi\,n)^{1/4}\Bigl(\dfrac{n}{e}\Bigr)^{n/2}}{2^n}=(2\pi\,n)^{1/4}\Bigl(\frac{\sqrt n}{2\sqrt e}\Bigr)^{n}. $$ Since $\dfrac{\sqrt n}{2\sqrt e}$ and $n^{1/4}$ converge to $\infty$, so does $\dfrac{\sqrt{n!}}{2^n}$.

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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.

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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

$$a_n=\frac{2^n}{\sqrt{n!}}\xrightarrow [n\to\infty]{} 0\Longrightarrow \frac{\sqrt{n!}}{2^n}\xrightarrow [n\to\infty]{}\infty$$

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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 $n>4$.

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