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### Proof of Stirling's Approximation

```Date: 03/09/2006 at 00:02:28
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

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

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

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/
```
Associated Topics:
College Calculus
High School Calculus
High School Sequences, Series

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