Let X be a Binomial B(\frac{1}{2},n), where n=2m.
Let a(m,k) = \frac{4^m}{(\stackrel{2m}{m})}P(X = m + k).
Show that lim_{m->\infty}(a(m,k))^2 = e^{-k^2}.
So far, I've found that P(X = m+k) = (\stackrel{2m}{m+k}) \frac{1}{4^m}
Then, a(m,k)=\frac{m!m!}{(m+k)!(m-k)!}.
But I have no...