# Upper bound on sum of binomial coefficients

If have been trying to prove the following limit: $\displaystyle\lim_{n\to \infty}\dfrac{\binom{n}{n/2}+\cdots+\binom{n}{n/2+\sqrt{n}}}{2^n}=0$ using the chernoff bounds for $\displaystyle\sum_{i=0}^{n/2+\sqrt{n}}\binom{n}{i}$ and then I suppose that the sum first $n/2$ terms should be easy to calculate and subtract. But thus far I have had no success in proving this.

I've thought of using Stirling's formula for each binomial coefficient, but then I see that that may not lead to anything.

I would appreciate I you could give me a hint to start my proof.

Let $X_n$ be a symmetric binomial random variable, i.e. $X_n \sim \operatorname{B}(n,1/2)$: $$\mathbb{P}(X_n = k) = \frac{1}{2^n} \binom{n}{k} \times [ 0 \leqslant k \leqslant n]$$ The mean and the variance of $X_n$ is well known: $$\mathbb{E}(X_n) = \frac{n}{2} \qquad \mathbb{Var}(X_n) = \frac{n}{4}$$ By the central limit theorem, the random variable $Z_n$ defined as $$Z_n = \frac{X_n - n/2}{\sqrt{n/4}} = \frac{2 X_n - n}{\sqrt{n}}$$ converges in distribution to the standard normal random variable. Therefore:
$$\sum_{n/2 \leqslant k \leqslant n/2 +\sqrt{n}} \binom{n}{k} \frac{1}{2^n} = \mathbb{P}\left( \frac{n}{2} \leqslant X_n \leqslant \frac{n}{2} + \sqrt{n} \right) = \mathbb{P}\left( 0 \leqslant \frac{2 X_n - n}{\sqrt{n} } \leqslant 2 \right) \stackrel{n\to \infty}{\longrightarrow} \Phi(2) - \Phi(0)$$ With $\Phi(0) = \frac{1}{2}$ this is exactly what Robert stated in his answer.
By the De Moivre-Laplace theorem, the limit is not zero: it is $\Phi(2) - 1/2$, where $\Phi$ is the standard normal cumulative distribution function.