For a Binomial$(n,p)$ random variable $X$, I'm interested in showing that $$ \frac{P(X>c)}{P(X>c-1)}=1-o(1) $$ uniformly in $c\in\mathcal{R}$, where $\mathcal{R}$ is the range of interest (Note that $c$ will vary with $n$). The $o(1)$ rate is meant as $n\to\infty$.

Now I have the following results (Note that $q=1-p$):

###Result 1 For $0\leq k\leq n$, set $$ P(X=k)=\frac{1}{\sqrt{2\pi pq n}}\exp\left(-\frac{(k-np)^2}{2npq} \right)(1+\delta_n(k)) $$ Then for every positive real sequence $\{c_n\}$ approaching zero, $$ \lim_{n\to\infty}\max_{k:|k-np|<c_n n^{2/3}}|\delta_n(k)|=0 $$

###Result 2 Suppose that $\{a_n\}$ is a sequence of real numbers such that $\lim_{n\to\infty}a_n=+\infty$ and $\lim_{n\to\infty}a_n n^{-1/6}=0$. Then $$ P(X\geq np+a_n\sqrt{npq})\sim \frac{1}{a_n\sqrt{2\pi}}\exp(-a_n^2/2) $$ where "$\sim$" means asymptotic equivalence.

Now, \begin{align} \frac{P(X>c)}{P(X>c-1)}&=\frac{P(X>c-1)-P(X=c)}{P(X>c-1)}\\ &=1-\frac{P(X=c)}{P(X\geq c)} \end{align}

Now, it is clear from both results that the upper bound in $\mathcal{R}$ should be $np+o(n^{2/3})$. What would the lower bound be? It seems that it is $np+n^{1/2+\gamma}$ for some very small $\gamma<1/6$ (e.g. $np+\sqrt{n\log\log n}$). Is this correct?

EDIT I tried some simple numerical computations and it seems that as long as $|c-np|= o(n^{2/3})$, the desired result holds. This suggests that a lower bound for $c$ could be $np$. Could someone verify?