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}

**EDIT**
From the the above results, the range $\mathcal{R}$ can be at least $np$ and at most $np+c_{n}\sqrt{npq}$, for some $c_n=o(n^{1/6})$. I can certainly show that at the extremes of the range, the ratio is $o(1)$. However, I think I also need to show that either **a)** for some value $\tilde{c}$ in between the extremes, $P(X=\tilde{c})/P(X\geq \tilde{c})=o(1)$ or that **b)** the ratio itself is monotonic (based on some numerical experiments, I think it is increasing in $c$). I've tried to go through the route of **b)** and show that $P(X=c)/P(X\geq c)\leq P(X=c+1)/P(X\geq c+1)$, but can't seem to get the math to work out.

**EDIT2**
Going the **a)** route, could someone comment if this is rigorous enough:
Suppose $\tilde{c}=np+c^*$, where $0<c^*\leq o(n^{1/6})$. Then 
\begin{align*}
\frac{P(X=\tilde{c})}{P(X\geq\tilde{c})}&=\frac{P(X=np+c^*)}{P(X\geq np+c^*)}\\
&=\frac{(npq)^{-1/2}\phi(c^*/\sqrt{npq})(1+o(1))}{\phi(c^*/\sqrt{npq})/(c^*/\sqrt{npq})}\\
&=\frac{c^*}{npq}=o(1)
\end{align*}