How much does continuity correction improve the normal approximation of binomial? From the central limit theorem, we know that $(\text{Bin}(n,p)-np)/\sqrt{npq}$ is approximately normal, with $q=1-p$. The Berry-Esseen theorem gives an upper bound for the error of this approximation: 
$$
\sup_{x\in \mathbb R}\left|\mathbb P\left(\frac{\text{Bin}(n,p)-np}{\sqrt{npq}}\le x\right)-\Phi(x)\right|\le \frac{C(p^2+q^2)}{\sqrt{npq}}\le \frac{1}{\sqrt{npq}}\tag{$*$}
$$
Furthermore, this estimate is optimal up to a constant, because the largest point probability for $\text{Bin}(n,p)$ is on the order of $1/\sqrt{2\pi npq}$. 
However, it is common practice to use a "continutity correction" when approximating a binomial distribtuion by a normal. I was wondering whether this correction leads to significant improvements in the error. That is, are there any convenient upper bounds for 
$$
\sup_{k\in \mathbb Z} \left|\mathbb P\left(\text{Bin}(n,p)\le k\right)-\Phi\left(\frac{k+\frac12-np}{\sqrt{npq}}\right)\right|
$$
which are better than ($*$)?

$\def\Bin{\text{Bin}}$Ok, I found the answer. Except in the case $p=1/2$, the continuity correction is not enough to improve the Berry-Esseen bound. 
Precisely, letting $F(k)=P(\Bin(n,p)\le k)$, and $z=(k+\frac12-np)/\sqrt{np(1-p)}$, and letting $\phi(z)$ be the pdf of the standard normal, then
$$
F(k) = \Phi(z) +\frac{p-q}{6\sqrt{npq}}(z^2-1)\phi(z)+O(1/n)
$$
Only when $p=q$ does the $1/\sqrt{n}$ term disappear.
