A question about limit $\lim_{n\to\infty} {n \choose n/2}/2^n$ My question is: What is the result of this limit: 
$\displaystyle \lim_{n \to +\infty} \frac{{n \choose n/2}}{2^n}=$ ?

As $\lim_{n\to+\infty} \frac{1}{2^n} \binom{n}{n/2} = \lim_{n\to+\infty} \frac{1}{4^n} \binom{2n}{n}$, and since for $r(n) = \frac{1}{4^n} \binom{2n}{n}$ the ratio
$$
r(n) >0 \land \frac{r(n+1)}{r(n)} = \frac{2n+1}{2(n+1)} < 1   \qquad \forall n \ge 1 
$$
it follows that $\lim_{n \to \infty} r(n) = 0$ since $r(n)$ monotonically decreases as $n$ grows.
Added: To address the gap in my demonstration pointed out by Henning:
$$
 \begin{eqnarray}
 r(n) = \frac{1}{4^n} \binom{2n}{n} &=& r(0) \prod_{k=0}^{n-1} \frac{r(k+1)}{r(k)} = \prod_{k=0}^{n-1} \frac{2k+1}{2(k+1)} \\ 
  &=& \prod_{k=0}^{n-1}  \left( 1 - \frac{1}{2(k+1)} \right) \le \prod_{k=0}^{n-1}  \left( 1 + \frac{1}{2(k+1)} \right)^{-1} \\
 &\le& \frac{1}{\sum_{k=0}^{n-1} \frac{1}{2(k+1)}}  = \frac{2}{H_n}
 \end{eqnarray}
$$
Thus $0 < r(n) \le \frac{2}{H_n}$ and $\lim_{n\to \infty} r(n) = 0$ follows from $\lim_{n\to\infty} \frac{2}{H_n} = 0$.
