I've got a binomial distributed random variable X where $p = \frac{1}{2}$, so that $P(X \leq \frac{n}{2} + o(\sqrt{n})) = \sum_{i=0}^{\frac{n}{2}+o(\sqrt{n})}\binom{n}{i}(\frac{1}{2})^n$ and the standard deviation is $\frac{\sqrt{n}}{2}$.
Now it says that the normal approximation to the binomial distribution yields that this probability is $\frac{1}{2}+o(1)$.
I'm wondering how this approximation works in detail. Anyone who could explain how to come up with the latter probability?
The central limit theorem tells us that if $X_n\sim\mbox{Bin}(n,1/2)$ then for each fixed $x$ we have $$\lim_{n\to\infty} P\left( \frac{ X_n-n/2}{\sqrt{n/4}}\le x\right) = \Phi(x),$$ where we make use of the facts that $EX_n=n/2$ and $\mbox{SD}(X_n)=\sqrt{(1/2)(1/2)n}$.
As originally phrased the question asks for a proof that $$P(X_n\le \frac n 2 + o(\sqrt n) = \frac 1 2 + o(1).$$ This is the same[see footnote below] as asking if, whenever $e_n$ is such that $\lim_{n\to\infty}e_n=0$, it follows that $$\lim_{n\to\infty}P(X_n\le n/2+e_n\sqrt n)=1/2.$$ That is, if $e_n\to0$ implies $P(X_n\le n/2+e_n\sqrt n)\to1/2$.
You are interested in the event $A_n = [X_n\le n/2+e_n\sqrt n]$. For each $\epsilon>0$ we know $|e_n \sqrt n|<\epsilon\sqrt n/2$ for all $n$ sufficiently large. For such $n$ we have the following containment of events: $$ [X_n\le n/2-\epsilon\sqrt n/2]\subseteq A_n\subseteq [X_n\le n/2+\epsilon\sqrt n/2].$$ The CLT, applied to the left-hand event and right-hand events gives $$\Phi(-\epsilon)\le\liminf_{n\to\infty}P(A_n)$$ and $$\limsup_{n\to\infty}P(A_n)\le \Phi(\epsilon).$$
Since $\epsilon>0$ was arbitrary, and $\Phi$ is continuous at $0$, this implies $$\lim_{n\to\infty}P(A_n)=\Phi(0)=\frac 1 2,$$ which is another way of saying $P(A_n)=1/2+o(1).$
Note: recall, from here or a_n=here that to say $a_n=o(b_n)$ is to say (for nonzero $b_n$, at least) that $a_n/b_n\to 0$; in the special case $b_n=1$ we have $a_n=o(1)$ is the same as $a_n/(1) \to 0$, that is, $a_n\to0$.