Finding the Limit of Binomial CDF Let $X\sim \mathrm{Binom}(n,p)$. Then its CDF is $$F_X(x)=\sum_{k=0}^{\lfloor x\rfloor}\binom{n}{k}p^k(1-p)^{n-k}.$$ I am trying to determine the limit of $F_X(x)$ with respect to $n$, with $p$ remain fixed: $$\lim_{n\to\infty}F_X(x)=\lim_{n\to\infty}\sum_{k=0}^{\lfloor x\rfloor}\binom{n}{k}p^k(1-p)^{n-k}.$$ The first thing that came up is de Moivre–Laplace theorem, that is $\frac{X-np}{\sqrt{np(1-p)}}\xrightarrow{D}N(0,1)$. Can I conclude from that fact then "$X\xrightarrow{D} N(np, np(1-p))$" (not really since $np$ explodes), so  $$\lim_{n\to\infty}F_X(x)=\lim_{n\to\infty}\left[\frac{1}{2}+\frac{1}{2}\mathscr{E}\mathrm{rf}\left(\frac{x-np}{\sqrt{2np(1-p)}}\right)\right]?$$ If not, how should I continue?
 A: The de Moivre–Laplace limit theorem involves relocation by subtraction of $np$ and rescaling by dividing by $\sqrt{np(1-p)}$.  If you just want the limit of the c.d.f. at a fixed value of $x$ without doing those things, then the limit is $0$.  For example, suppose $x=40$ and $p=0.2$ and $n=10000000000000000000$.  What would you expect the probability that $X\le x$ to be?
PS: Here's a simple way to prove the limit is $0$.  As to the rate of convergence, this method will be conservative.
Chebyshev's inequality says that if a random variable $Y$ has expected value $\mu$ and variance $\tau^2$ then
$$
\Pr\left( |Y-\mu| > k \right) \le \frac{\tau^2}{k^2}.
$$
You have $\operatorname{E}(X)=np$ and $\operatorname{var}(X)= np(1-p)$ so
\begin{align}
\Pr(X=x) \le \Pr(X < x+1)  &= \Pr\left( X-np < x+1-np \right) \le \Pr\left( |X-np|> |x+1-np| \right)  \\[10pt]
& \le \frac{np(1-p)}{(x+1-np)^2} \to 0 \text{ as } n\to\infty
\end{align}
since we have the square of $n$ in the denominator and the first power of $n$ in the numerator.
A: You are on the right track.
A naive search for
"normal approximation to the binomial distribution"
produces many useful links.
