Stirling’s approximation and Big O

Question

How to prove that $2n \choose n$ = $\frac{2^{2n}}{\sqrt{\pi n}}(1 + O(1/n))$ using Stirling’s approximation?

I know how to prove that $2n \choose n$ = $\frac{2^{2n}}{\sqrt{\pi n}}$ but I am having trouble with the prof of big O.

Any help Would be appreciated. Thanks

Answer 1

Use Stirling's theorem in the form $n! =\sqrt{2\pi n}(n/e)^n(1+O(1/n)) $. This will give you the $O(1/n)$ that you want.