# Math Help - Stirling's approximation

1. ## 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?

2. ## 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 $(n+1)\log (n+1)$ .

3. ## 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?

4. ## Re: Stirling's approximation

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