# Stirling's approximation

• July 19th 2011, 07:06 PM
eulcer
Stirling's approximation
Hi, I was just checking out the proof of this in William Feller's book, and it jumps from the arithmetic mean of (nlog(n)-n) and ((n+1)log(n+1)-n) to ((n+(1/2))log(n)-n). I'm not seeing it. The book is certainly a good read, but nonetheless I'm missing out. Any hints?
• July 20th 2011, 01:44 AM
FernandoRevilla
Re: Stirling's approximation
Prove that $L=\dfrac{1}{2}\displaystyle\lim_{n \to{+}\infty}{\frac{n\log n+(n+1)\log (n+1)-2n}{(n+1/2)\log n-n}}=\ldots=1$

Hint: Divide numerator and denominator by $(n+1)\log (n+1)$ .
• July 20th 2011, 05:14 AM
eulcer
Re: Stirling's approximation
Alright.
It's not equal, it's only an approximation of the approximation.
And this is where the ~ comes from then?
• July 20th 2011, 05:19 AM
FernandoRevilla
Re: Stirling's approximation
Quote:

Originally Posted by eulcer
And this is where the ~ comes from then?

Right.