Showing this asymptotic relation using Stirling's formula

Question

The following question is from 'Numerical Analysis for Statisticians.'

It asks to prove, using Stirling's formula, that $${2n-1 \choose n } \sim \frac{2^{2n-1}}{\sqrt{\pi n}}$$

But after applying Stirling's formula, and rearranging, I get $${2n-1 \choose n} \sim \frac{1}{\sqrt{\pi n}} \sqrt{\frac{n-\frac{1}{2}}{n-1}} \times \frac{(2n-1)^{2n-1}}{n^n (n-1)^{n-1}}$$

So that to complete the question I need to show

$$\frac{(2n-1)^{2n-1}}{n^n (n-1)^{n-1}} \sim 2^{2n-1}$$

But not only am I unsure how to show this, playing around in R I'm also unsure if it's even true. What am I missing here?

Answer 1

$$A=\binom{2 n-1}{n}=\frac{(2 n-1)!}{(n-1)!\, n!}=\frac{\Gamma (2 n)}{\Gamma (n) \Gamma (n+1)}$$ $$\log(A)=\log(\Gamma (2 n))-\log(\Gamma (n))-\Gamma (n+1)$$

Now,use Stirling approximation $$\log(\Gamma(p))=p (\log (p)-1)+\frac{1}{2} \left(-\log \left({p}\right)+\log (2 \pi )\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$ Apply it three times and continue with Taylor expansions to get $$\log(A)=2 n \log (2)+\left(-\frac{1}{2} \log \left({n}\right)+\log \left(\frac{1}{2 \sqrt{\pi }}\right)\right)-\frac 1 {8n}+O\left(\frac{1}{n^2}\right)$$ $$A= e^{\log(A)}=\frac{2^{2 n-1}} {\sqrt{\pi n}}e^{-\frac 1 {8n}}\sim\frac{2^{2 n-1}} {\sqrt{\pi n}}$$

Answer 2

$$\frac{(2n-1)^{2n-1}}{n^n (n-1)^{n-1}}=2^{2n-1}\left(1-\frac{1}{2n}\right)^{-1} \left(1-\frac{1}{2n}\right)^{2n}\left(1-\frac{1}{n}\right)^{-n}.$$The last two factors converge to $e$ and $e^{-1}$ respectively, so the whole thing is asymptotic to $2^{2n-1}$.