Show that for a very large $\displaystyle m$, $\displaystyle C(m) \approx \frac{4^m}{m^{\frac{3}{2}} \times \sqrt{\pi}}

$

where $\displaystyle C(m)$ denotes the mth catalan number.

So I did:

$\displaystyle C(m) = \frac{1}{m+1} \binom{2m}{m} = \frac{(2m)!}{(m+1)!m!} = \frac{(2m)(2m-1)(2m-2)}{m!}$ after some canceling.

Now $\displaystyle \lim_{m \to \infty} \frac{(2m)(2m-1)(2m-2)}{m!}$ and I'm stuck here... we are given Stirling's approximation formula $\displaystyle n! \approx \sqrt{2\pi n} \times \left(\frac{n}{e}\right)^n$ for $\displaystyle n >>>>1$ but I do not know how to apply that here to my final step...

Any help would be greatly appreciated!

Thanks heaps!