Proving Stirling's Formula for Scientists

[Treadstone 71]

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

 

[Hurkyl]

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

 

[Treadstone 71]

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.

 

[Hurkyl]

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.

 

[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

 

[Treadstone 71]

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}

 

[Hurkyl]

The first one is fairly easy. It's just

<br /> \frac{n!}{n^n} = \frac{1}{n} \frac{2}{n} \cdots \frac{n}{n}<br />

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!)

 

[shmoe]

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

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).

 

[Treadstone 71]

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.

 

[shmoe]

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

 

[Treadstone 71]

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

 

[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 don't know how to do that...

 

[shmoe]

the variable e or exponential function can be found by lim (n --> infinity)1/n!

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.

 

[benorin]

\boxed{\mbox{This is analysis -- you don&#039;t need to be careful and precise -- you just need to make sure your errors go to zero!}}​

Nice

 

[benorin]

There's a latex tutorial around somewhere as well.

Indeed there is: PF resources for learning LaTeX Math Typesetting (a recent post of mine in which I had gone and dug it up.)

 

[benorin]

A direct proof of your limit...

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

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 argument 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}}​

 

[Treadstone 71]

Thanks. It will take me some time to digest that.

 

[devious_]

Very nice ben!