You're right. It is easy with [Stirling's approximation][1]
$$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$
so that 
$$\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\pi\sqrt{2}N}\rightarrow 0$$
as $N\rightarrow\infty$.

[1]: https://en.wikipedia.org/wiki/Stirling's_approximation