Proof of Stirling's ApproximationDate: 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/ |
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