# Help with Stirling's formula

1. Mar 9, 2006

How do I show that $$\frac{(n!)^{1/n}}{n}\rightarrow \frac{1}{e}$$?

Last edited by a moderator: Jul 15, 2014
2. Mar 9, 2006

### Hurkyl

Staff Emeritus
Stirling's formula is almost always a good idea when dealing with factorials and asymptotics.

3. Mar 9, 2006

I looked through Stirling's formula, and it only gives approximations. I'm not sure those approximations are valid at limits, as when I plug it in the original equation, I get 0 as the limit instead of 1/e.

4. Mar 9, 2006

### Hurkyl

Staff Emeritus
Yes, but (good) approximations have bounds, or some other theoretical fact for using them rigorously. For example, consider the inequalities near the bottom of mathworld's page.

And, I think, you have the theorem that (n!) / (Stirling's formula) --> 1 as n --> infinity.

By the way, could you show your work? I think you made a mistake in your arithmetic... it seems to work for me.

5. Mar 9, 2006

### shmoe

The mathworld page is a little unclear on what the double ~ means. To bolster Hurkyl's "I think", I will say they are most definitely meaning they are asymptotic, you have:

$$\lim_{n\rightarrow\infty}\frac{n!}{n^{n+1/2}e^{-n}\sqrt{2\pi}}=1$$

6. Mar 9, 2006

I re-checked my arithmetic and it works. However, I don't think I can use Stirling's formula, since we have not seen it yet. This is supposed to be an analysis 2 assignment. Is there some other way to show that

$$\frac{n!}{n^n}\rightarrow 0$$

or

$$\frac{(n!)^{1/n}}{n}\rightarrow \frac{1}{e}$$

7. Mar 9, 2006

### Hurkyl

Staff Emeritus
The first one is fairly easy. It's just

$$\frac{n!}{n^n} = \frac{1}{n} \frac{2}{n} \cdots \frac{n}{n}$$

which can be grouped in a convenient fashion.

(This is analysis -- you don't need to be careful and precise -- you just need to make sure your errors go to zero!)

8. Mar 9, 2006

### shmoe

This can be done in an very elementary way, what have you tried?

My first thought for the second without invoking stirlings is to essentially mimic the proof of stirlings, it's not terribly hard (rough details on the mathworld page).

9. Mar 9, 2006

I got the first part. The second part, instead of showing that it converges to 1/e, what if I want to show that it converges to a number <1? Again, without evoking Stirling's.

10. Mar 9, 2006

### shmoe

You should be able to get <=1 by using your proof of the first result.

11. Mar 9, 2006

I got it. I was hoping to use some kind of squeeze theorem to get it to 1/e. Thanks for the help.

12. Mar 9, 2006

### kahless2005

the variable e or exponential function can be found by lim (n --> infinity)1/n! Therefore n!/n, with the same limit, will give you 1/e. Its a rather simple Power Function.

Sorry about the lack of nice mathematical appeartance.. I dont know how to do that...

13. Mar 9, 2006

### shmoe

Are you referring to the infinite series

$$\sum_{n=0}^{\infty}\frac{1}{n!}=e$$

I don't see how that will help, but I can't think of what else you could mean.

If you click on the above, you can see how it was made. There's a latex tutorial around somewhere as well.

14. Mar 10, 2006

### benorin

Nice :rofl:

15. Mar 10, 2006

### benorin

16. Mar 10, 2006

### benorin

A direct proof of your limit...

I WILL GIVE A DIRECT PROOF (without using Stirling's Formula):

Let the limit be denoted Y so that

$$Y= \lim_{n\rightarrow\infty} \frac{(n!)^{\frac{1}{n}}}{n} = \lim_{n\rightarrow\infty} e^{\displaystyle{\log \left( \frac{(n!)^{\frac{1}{n}}}{n}\right)}} = \lim_{n\rightarrow\infty} e^{\displaystyle{\frac{1}{n}\log \left( \frac{n!}{n^n}\right)}} = \lim_{n\rightarrow\infty} e^{\displaystyle{\frac{1}{n}\log \left( \prod_{k=1}^{n} \frac{k}{n}\right)}} = \lim_{n\rightarrow\infty} e^{\displaystyle{\frac{1}{n}\sum_{k=1}^{n}\log \left( \frac{k}{n}\right)}}$$
$$\mbox{ } = e^{\displaystyle{\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\log \left( \frac{k}{n}\right)}} = e^{\displaystyle{\lim_{r\rightarrow 0^+} \int_{r}^{1} \log (x)\,dx}} ,$$

where the limit was passed through to the arguement of the exponential function by reason of continuity and the sum was seen to be a Riemann sum of the form $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k=1}^{n}f \left( \frac{k}{n}\right) = \int_{0}^{1}f(x)\, dx,$ which is improper integral in this case becase of the discontinuity of log(x) at x=0; the limit becomes

$$Y =\displaystyle{ e^{\displaystyle{\lim_{r\rightarrow 0^+} \int_{r}^{1} \log (x)\, dx }}= e^{\displaystyle{\lim_{r\rightarrow 0^+} \left[ x(\log (x)-1)\right]_{x=r}^{1}}} = e^{\displaystyle{\lim_{r\rightarrow 0^+} \left[ -1-r(\log (r)-1)\right]}} = e^{-1}e^{\displaystyle{\lim_{r\rightarrow 0^+} \frac{1-\log r}{r^{-1}}}} =^{H} e^{-1}e^{\displaystyle{\lim_{r\rightarrow 0^+} \frac{-r^{-1}}{-r^{-2}}}} =e^{-1}e^{\displaystyle{\lim_{r\rightarrow 0^+} r}} = e^{-1}}$$

where $$=^{H}$$ denoted the use of l'Hospital's rule and whereby we may conclude that

$$\boxed{ \lim_{n\rightarrow\infty} \frac{(n!)^{1/n}}{n}= \frac{1}{e}}$$​

Last edited: Mar 10, 2006
17. Mar 10, 2006