2
$\begingroup$

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.

$\endgroup$
13
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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 $n>4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.