Determine $\lim\limits_{n \to \infty}{{n} \choose {\frac{n}{2}}}\frac{1}{2^n}$, where each $n$ is even For each positive even integer $n$, set 
$$P_n = \displaystyle {{n} \choose {\frac{n}{2}}}\frac{1}{2^n}.$$
Show that $\displaystyle \lim_{n \to \infty} P_n$ exists and determine its value. 
Here's what I have so far: each $P_n$ can be thought of as the probability of tossing a fair coin $n$ times, and obtaining exactly $\frac{n}{2}$ heads. My expectation is that $P_n \to 1$, since in any large trial, one would expect to record just about as many heads as tails. But I'm not sure how to mathematically justify this hunch.
Hints or solutions are greatly appreciated.

Here's another approach! Write $n=2m:$
$$\binom{2m}{m}\frac{1}{4^m}=\frac{1}{\pi}\int_0^1 t^{-\frac{1}{2}}(1-t)^{m-\frac{1}{2}}dt$$
The integrand tends to $0$ as $m\to \infty$
