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I came across this limit in some context:

$$ \lim\limits_{n\rightarrow \infty} (n!)^{\frac{1}{n}}$$

I could only say that $n! > n$ implies the limit is greater than or equal to $1$. However, the result seems to be infinity. I do not know how to arrive at this result.

Any ideas?

Based on the answer and the comment below, I wonder if it is possible to prove this using elementary Calculus?

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marked as duplicate by Lord_Farin, Dan Rust, MathOverview, Andrey Rekalo, Thomas Jul 6 '13 at 18:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Use Stirling's formula : $n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$ – zuggg Jul 6 '13 at 6:46
Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – Lord_Farin Jul 6 '13 at 7:33
Here is a related, if not exact, question with many nice answers. – David Mitra Jul 6 '13 at 10:31
@Arjang I am a bit annoyed to have to point you to this again: do not remove all non-TeX entities from titles! – Lord_Farin Jul 6 '13 at 11:21
@Arjang MSE is not a maths magazine. Titles should be as descriptive as possible; AFAIC, brevity is at best secondary and is overridden by clarity considerations. Perhaps I overreacted, but of course I can hardly go check all your edits to see if this is only the second occurrence; this all notwithstanding, your efforts are appreciated. :) – Lord_Farin Jul 6 '13 at 17:35
up vote 4 down vote accepted

You can do it like this. Fix an arbitrary $n_0$ and see what happens when $n > n_0$.

Clearly, $$ n! = n_0! \cdot (n_0+1) (n_0+2) \ldots (n). $$ It follows that $$ n! \geq n_0! \cdot n_0^{n - n_0}, $$ and $$ \sqrt[n]{n!} \geq \sqrt[n]{n_0! \cdot n_0^{n-n_0}} = \sqrt[n]{\frac{n_0!}{n_0^{n_0}}} \cdot n_0. $$ Now let's see what happens when $n \to \infty$. The right hand side converges to $n_0$. So, when $n$ is large enough, we can say, for example, that $\sqrt[n]{n!} \geq \frac{1}{2}n_0$.

But, since $n_0$ is arbitrary, we see that for any positive $C$ we will have $\sqrt[n]{n!} \geq C$ for all large enough $n$. It means that $\sqrt[n]{n!}$ tends to infinity.

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The series $\displaystyle \sum_{n= 0}^\infty\frac{x^n}{n!}$ converges for all $x$, so $\displaystyle \lim_{n\to\infty}\frac{x^n}{n!}=0$ for any $x$. This means in particular that, given any fixed $M$, we have $n!>M^n$ for $n$ large enough. Thus $(n!)^{1/n}>M$ for $n$ large enough. Since $M$ was arbitrary, $\lim_{n\to\infty}(n!)^{1/n}=\infty$.

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That is a lovely argument – Vishal Gupta Jul 6 '13 at 7:13
Yes, it's lovely indeed. – user1551 Jul 6 '13 at 10:23
@Andres Caicedo We could prove that $\lim\limits_{n\to \infty}\frac{x^{n}}{n!} = 0$ independently of the convergence of series though. – Vishal Gupta Jul 8 '13 at 3:30
Yes, of course, and several of the arguments here establish that (implicitly) along the way. – Andrés E. Caicedo Jul 8 '13 at 4:12

The most elementary proof I know goes like this: note that $n! \ge n(n-1)\cdots \frac n2 \ge (\frac n2)^{n/2}$. (The middle step needs fixing if $n$ is odd, but the inequality $n! \ge (\frac n2)^{n/2}$ still stands.) Therefore $(n!)^{1/n} \ge (\frac n2)^{1/2}$, which tends to infinity; hence $(n!)^{1/n}$ itself tends to infinity.

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use this well kown $$\left(\dfrac{n}{e}\right)^n<n!<e\left(\dfrac{n}{2}\right)^n$$

or you can use $$n!\approx\left(\dfrac{n}{e}\right)^n\sqrt{2n\pi}$$

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Thanks. Could you provide some reference for the well known result? – Vishal Gupta Jul 6 '13 at 6:55

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