Finding the limit as $k$ tends to infinity of this sum $\sum_{i=0}^{\lceil zk \rceil}{k\choose i} p^i(1-p)^{k-i}$
$z, p \in [0,1]$ 
I am looking to find the limit as $k$ tends to infinity but don't know how I would do this
 A: Just a possible hint, too long for a comment. 
The cumulative distribution function for the binomial distribution $P(s; k, p)$ (probability of $s$ or less successes over $k$ trials, each with a probability $p$ of success), is given by $$ P(s; k, p) = \sum_{i=0}^{\lfloor s \rfloor}  = {k \choose i} p^{i} (1-p)^{k-1}$$
Your expression has then a straightforward interpretation: the probability of no more than $\lfloor zk \rfloor$ successes over $k$ trials. One then considers the fact that the cumulative binomial tends to a normal distribution with mean $kp$ and variance $kp(1-p)$ in the limit $k \to \infty$. Hope it is of some help.
A: If $\{X_i\}$ is an iid sequence of Bernoulli random variables with $P(X_i=1)=p$ and $S_k=\sum_{i=1}^kX_i$, then
$$\sum_{i=0}^{\lceil zk\rceil}{k\choose i}p^k(1-p)^{k-i}=P(S_k\le\lceil zk\rceil).$$
By the law of large numbers, $S_k/k\to p$ (almost surely or in probability). This implies that the limit is zero if $z<p$ and one if $z>p$. The case $z=p$ is more delicate and may require a local central limit theorem.
