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.



Showing $o(1)$ convergence for ratio of successive binomial tail probabilities
I think, if $c=np+o(n)$, you may simply use $P(X=c+1)=P(X=c)(1+o(1))$, and so
$P(X\in \{c,c+1,\dots,c+M-1\})=(M+o(1))P(X=c)$ for any fixed $M$.


