Hi all, I was wondering if someone could help me show

$\displaystyle \lim_{n\rightarrow\infty}{2n \choose n}\left(\frac{1}{4}\right)^n=0$

I was able to do some algebra to show that this limit is

$\displaystyle \lim_{n\rightarrow\infty} \frac{1}{2}\frac{3}{4}\frac{5}{6}\dots \frac{2n-1}{2n}$

which must converge since the sequence is decreasing and bounded by 0. But I haven't been able to figure out how to show the limit is actually 0 using the epsilon/delta definition of convergence. Is there an alternative way that I might prove the limit is 0, or can someone give me a hint with regard to the epsilon/delta definition?

Thanks