Limit of $p_{k+1}(z)/p_k(z)$ with $p_k(z)=\sum\limits_{i=0}^{\lceil zk \rceil}{k\choose i} p^i(1-p)^{k-i}$ and $z
How to find the limit of $\dfrac{p_{k+1}(z)}{p_k(z)}$ as $k$ tends to infinity, where, for every $k$, $$p_k(z)=\sum_{i=0}^{\lceil zk \rceil}{k\choose i} p^i(1-p)^{k-i}$$ for some $z$ and $p$ in $(0,1)$ such that $z<p$?
The top and bottom both converge to $0$ by the law of large numbers. 
I was thinking of using L'Hopital with respect to $k$ but not sure how that would work because of the ceil function. 
 A: Let $(X_n)$ be i.i.d. Bernoulli with parameter $p$, then $$p_k(z)=P(X_1+\cdots+X_k\leqslant kz)$$
Since $z<p$, the events $[X_1+\cdots+X_k\leqslant kz]$ are large deviations events and one knows that $$p_k(z)=e^{-kI(z)+o(k)}$$ 
where $I$ denotes the so-called action of the large deviations principle, defined in general as $$I(z)=\sup_{t<0}\left(tz-\log E(e^{tX})\right)$$
In the present case, $$E(e^{tX})=pe^t+1-p$$ hence $$I(z)=z\log\left(\frac{z}p\right)+(1-z)\log\left(\frac{1-z}{1-p}\right)$$
This suggests that $p_{k+1}(z)/p_k(z)$ should converge to $e^{-I(z)}$, but, to conclude this, one should refine the $e^{o(k)}$ error term in the asymptotic expansion of $p_k(z)$, to $$\frac{c(z)+o(1)}{\sqrt{k}}$$ for some positive $c(z)$ whose value will be irrelevant. And guess what, it happens that this holds true, since $$p_k(z)\sim\frac{e^{-kI(z)}}{\sqrt{2\pi z(1-z)}}$$ in the precise sense that $$\lim_{k\to\infty}\sqrt{2\pi z(1-z)}\,e^{kI(z)}p_k(z)=1$$ hence indeed, $$\lim_{k\to\infty}\frac{p_{k+1}(z)}{p_k(z)}=e^{-I(z)}=\left(\frac{p}z\right)^z\left(\frac{1-p}{1-z}\right)^{1-z}$$
A: If $X\sim\operatorname{Bin}(n,p)$, then $\frac{X-np}{\sqrt{np(1-p)}} \xrightarrow{d} N(0,1)$ and one can use it to estimate numerator and denominator. Basically, both are $\Phi(\frac{\sqrt{n}(z-p)}{\sqrt{p(1-p)}})(1+O(\frac1n))$, so the limit is 1.
