# Difficult limit formula

1. Dec 2, 2008

### jostpuur

I have some reasons to believe that this equation is true:

$$\lim_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \frac{(2n)!}{(n!)^2} = \frac{1}{\sqrt{\pi}}$$

Anyone having idea of the proof? I don't even know how to prove that the limit is strictly between zero and infinity.

2. Dec 2, 2008

### daudaudaudau

I can do it with Stirlings approximation.

3. Dec 2, 2008

### jostpuur

ok. Thank's for reminding of it.