Stirling on ${2n-1 \choose n}$ I'm trying to find an expression for 
$${2n-1 \choose n}$$ using Stirling's approximation
$$k!\sim \sqrt{2\pi k}(\frac{k}{e})^k.$$
I see
$${2n-1 \choose n}\approx \frac{1}{\sqrt{2\pi}}\sqrt{\frac{2n-1}{n(n-1)}}\frac{(2n-1)^{2n-1}}{n^n(n-1)^{n-1}}$$
but cannot get much further. I could do
\begin{align*}
{2n-1 \choose n}&\approx \frac{1}{(2n-1)}\frac{1}{\sqrt{2\pi}}\sqrt{\frac{2n-1}{n(n-1)}} \frac{(2n-1)^{n-1}}{n^n}\frac{(2n-1)^{n-1}}{(n-1)^{n-1}} \\
&= \frac{1}{(2n-1)}\frac{1}{n\sqrt{2\pi}}\sqrt{\frac{2n-1}{n(n-1)}}(\frac{2n-1}{n})^{n-1}(\frac{2n-1}{n-1})^{n-1}
\end{align*}
but I don't see how that should work. I have done the calculation before and used $(\frac{n-1}{n})^n$=e for large n but $2n-1$ does not really help for this.

It looks like it's asymptotic to $\dfrac1{\sqrt{\pi n}} 2^{2n-1}$. For example, write
$$\left(\frac{2n-1}n\right)^n\left(\frac{2n-1}{n-1}\right)^{n-1} = \left(\frac{2n-1}{2n}\right)^n\left(\frac{2(n-1)+1}{2(n-1)}\right)^{n-1}\cdot 2^n\cdot 2^{n-1}.$$
Then note 
$$\left(\frac{2n-1}{2n}\right)^n = \left(\big(1-\frac1{2n}\big)^{2n}\right)^{1/2} \sim e^{-1/2},$$
and so on.
