Infinite sum over binomial coefficients up to $\frac{1}{2}n+\frac{1}{2}\sqrt{n}$ 
I want to evaluate the limit $$L=\lim_{n \rightarrow \infty} \frac{1}{2^n} \times \sum_{k=0}^{\frac{1}{2}n+\frac{1}{2}\sqrt{n}}\binom{n}{k}.$$

Maybe some elementary identies involving the binomial coefficients can help, but the only one I know, which looks similar to mine, is $$\sum_{k=0}^{\frac{1}{2}n}\binom{n}{k}=2^{n-1} \Rightarrow \lim_{n \rightarrow \infty} \frac{1}{2^n} \times \sum_{k=0}^{\frac{1}{2}n}\binom{n}{k}=\frac{1}{2}$$
and that gives me $L \geq \frac{1}{2}$. Also I found a generalized version in an answer of this question on MO. 
In the answer it is showen that $L=g(\frac{1}{2})$, but I have no idea what the function $g$ is.


Thanks for your ideas, comments and answers!
 A: The function $g$ is $g(\alpha) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\alpha} e^{-x^{2}/2}dx$. The main idea is using the central limit theorem:
Let $X_i$ be a Bernoulli distribution with $p = 1/2$. Then $\mathbb{E}[X_i] = 1/2$ and $\sigma(X_i) = 1/2$, so we have
$$
\frac{\bar{X} - 1/2}{1/(2\sqrt{n})}\to Z\,(\text{standard normal distribution})
$$
in the sense that
$$
\mathbb{P}\left[\frac{\bar{X} - 1/2}{1/(2\sqrt{n})} \leq \alpha\right] \to \mathbb{P}[Z\leq \alpha] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\alpha}e^{-x^{2}/2}dx
$$
as $n\to\infty$. By the way, $X_1 + \cdots +X_n$ follows a binomial distribution with $n = n$ (sorry for the abusing notation) and $p = 1/2$, so $$
(LHS) = \mathbb{P}\left[X_{1} + \cdots + X_{n} \leq n\left(\frac{1}{2} + \frac{\alpha}{2\sqrt{n}}\right)\right] = \sum_{k=0}^{\frac{n}{2} + \frac{\alpha\sqrt{n}}{2}} \frac{1}{2^{n}}\binom{n}{k}
$$
and we get the result. Note that there's no closed form of $g(1)$ in terms of elementry functions. It is approximately 0.8413...
