Proof of Stirling's Approximation

Date: 03/09/2006 at 00:02:28
From: Guadalupe
Subject: Calculus sequence proof

I need the proof for the following:

  lim ((e^n)(n!)) / ((n^n)(n)^1/2  = (2pi)^1/2

Date: 03/09/2006 at 08:25:09
From: Doctor Vogler
Subject: Re: Calculus sequence proof

Hi Guadalupe,

Thanks for writing to Dr. Math.  It sounds to me like you are asking
for a proof of Stirling's Approximation, as mentioned in:

  Stirling's Approximation
    http://mathforum.org/library/drmath/view/55996.html 

The first proof that I saw of Stirling's Approximation used a summation

   n
  sum ln i
  i=1

which, as you'll notice, is the log of n!.  Consider the function

  f(x) = ln x

and use one of the inequalities listed in

  Formula to Sum a Series of Square Roots
    http://mathforum.org/library/drmath/view/65309.html 

These are proved by noticing things like: if f(x) is an increasing
function, then

  f(n) >= (integral from n-1 to n of f(x) dx) >= f(n-1).

Better yet, if f is concave down, then the trapezoid under f(x) has area

  (1/2)(f(n-1) + f(n))

which is less than the total area under f(x), which is

  integral from n-1 to n of f(x) dx.

Unfortunately, this only showed that the limit was between e/sqrt(2)
and e, and didn't give an exact value.  But you can find another proof at

  Nrich
    http://nrich.maths.org/askedNRICH/edited/2614.html 

which gives the exact value sqrt(2*pi).

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/