$\lim_{ n\to \infty} \frac{n!}{n^n}$ via L'Hospital's rule

Question

I just need to find this limit and I don't know how to use L'Hopital's rule in this case:

$$\lim_{ n\to \infty} \frac{n!}{n^n}.$$

I apologize for the lack of formatting, I've never used the site before.

look up Stirling's formula, it is divergent, the limit is $\infty$. I don't think one can calculate such limit using L'Hopital's rule. — Rogelio Molina, Mar 31 '15 at 06:34
You couldn't use L'Hopital because we're dealing with sequences, not real valued functions. Stirling's formula doesn't apply either. — , Mar 31 '15 at 06:35
I don't see why Stirling's formula wouldn't work, but it is overkill. — JessicaK, Mar 31 '15 at 06:39

score 9 · Answer 1 · answered Mar 31 '15 at 06:33

9

hint: $\dfrac{n!}{n^n} < \dfrac{1}{n}$

answered Mar 31 '15 at 06:33

DeepSea

77,651

That would seem to make things clear, but what process can I show as my work? I'm afraid I don't understand the forces at work here. – Logan Mar 31 '15 at 06:40
You can use the squeeze theorem here: $0 < \dfrac{n!}{n^n} < \dfrac{1}{n} \to$ limit $ = 0$. – DeepSea Mar 31 '15 at 06:41
Oh wow - okay, that does make a lot more sense now. Though is there something I need to do to confirm that $\frac{n!}{n^n}$ < $\frac{1}{n}$ ? – Logan Mar 31 '15 at 06:47
1

$\dfrac{n!}{n^n} = \dfrac{1\cdot 2\cdots n}{n\cdot n\cdots n} < \dfrac{1}{n}\cdot 1\cdot 1\cdots 1 = \dfrac{1}{n}$ – DeepSea Mar 31 '15 at 06:49

score 2 · Answer 2 · answered Mar 31 '15 at 06:41

2

Consider the series $\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}$ and set $a_n=\frac{n!}{n^n}$. By ratio test $\lim\limits_{n->\infty}\frac{a_{n+1}}{a_n}=\frac{1}{e}<1$. Hence this series is convergent. It follows that $\lim\limits_{n->\infty}\frac{n!}{n^n}=0$.

answered Mar 31 '15 at 06:41

math

949

What do you think about the value of that infinite series? – Bumblebee Mar 31 '15 at 07:31

score 0 · Answer 3 · answered Mar 31 '15 at 07:03

You can use that $n!$'s order of magnitude is lower than $n^n$'s, so if you are watching $\frac{n!}{n^n}$, the limes must be $0$.

The same way you can prove for example, that $\frac {logn}{n}$ has a limit of $0$ too.

Here are the order of magnitude, in order:

$log n << \sqrt n << n << n^x << x^n << n! << n^n$

All of these are going towars $\infty$, but you can see that left to the right they are faster and faster.

If you divide a smaller one with a bigger one, the limit is $0$, and if you divide a bigger one with a smaller one, the limit is $\infty$

Hope this helps a bit. :)

Bumblebee · Answer 4 · 2015-03-31T07:27:04.803

Consider the Gamma Function $$\Gamma(n+1)=\int_0^{\infty}x^ne^{-x}dx$$ For integer values of $n$ we have $\Gamma(n+1)=n!$

Also for large $n$ values, we can use The Stirling's Approximation $$\Gamma(n+1)\sim\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n$$ Therefore $$\dfrac{\Gamma(n+1)}{n^n}\sim\sqrt{2\pi n}\left(\dfrac{1}{e}\right)^n$$ Using L'Hospital's Rule we can show that $$\lim_{n\to\infty}\dfrac{\sqrt n}{e^n}=0$$ Hence $$\lim_{n\to\infty}\dfrac{n!}{n^n}=0.$$

$\lim_{ n\to \infty} \frac{n!}{n^n}$ via L'Hospital's rule

4 Answers4