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.
 A: Stirling's approximation gives the following asymptotics for the central binomial coefficient:
$$
{2n \choose n} \sim \frac{4^n}{\sqrt{\pi n}}\text{ as }n\rightarrow\infty
$$
Therefore,
$$
{2n-1 \choose n}
= \frac{n}{2n}{2n \choose n}
\sim \frac{4^n}{2\sqrt{\pi n}}
$$
A: 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.
A: Continuing from OP:
$\quad\displaystyle\frac{1}{(2n-1)}\frac{1}{n\sqrt{2\pi}}\sqrt{\frac{2n-1}{n(n-1)}}\left(\frac{2n-1}{n}\right)^{n-1}\left(\frac{2n-1}{n-1}\right)^{n-1}$
$=\displaystyle\frac{2^{2n}}{(2n-1)}\frac{1}{n\sqrt{2\pi}}\sqrt{\frac{2n-1}{n(n-1)}}\left(1+\frac{-0.5}{n}\right)^{n-1}\left(1+\frac{0.5}{n-1}\right)^{n-1}$
$\approx\displaystyle\frac{2^{2n}}{(2n-1)}\frac{1}{n\sqrt{2\pi}}\sqrt{\frac{2n-1}{n(n-1)}}e^{-0.5}e^{0.5}\left(\frac{n}{n-0.5}\right)$
$=\displaystyle\frac{2^{2n+1}}{(2n-1)}\frac{1}{\sqrt{2\pi}}\sqrt{\frac{1}{n(n-1)}}$
