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. 

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.
