proof for distribtion of proportions of a binomial variable Is it true that the proportions of a binomial variable:
$$  {\displaystyle f(k,n,p)=\Pr(k;n,p)=\Pr(X=k)={\binom {n}{k}}p^{k}(1-p)^{n-k}} $$ 
follows a normal distribtion when n approaches infinity with 
$$ \mu=p  $$ 
and
$$ \sigma=\sqrt{\frac{p(1-p)}{n}} $$ 
if that is true, how to prove it?
 A: What you wrote is not literally true, but it is correct in spirit. For given $n$ and $p$, your function
$$
\binom{n}{k}p^k(1-p)^{n-k}
$$
is the probability mass function of a random variable $S_n$ with mean $np$ and variance $np(1-p)$. In fact, this follows from the equality in distribution
$$
S_n\overset{d}{=}X_1+\cdots+X_n,
$$
(where $X_1,\ldots,X_n$ is a sequence of independent Bernoulli random variables) together with the additivity of mean and variance for independent random variables.
The strong law of large numbers tells us that $S_n$ will concentrate around $np$, meaning that all $k$ values with $|k-np|$ more than a constant times $n$ will be inconsequential in the $n\to\infty$ limit. The central limit theorem tells us exactly what scaling we need to make the limit interesting:
$$
k=np+x\sqrt{np(1-p)}.
$$
So we are looking at $k$ that is exactly $x$ standard deviations away from the mean, $np$. Then
$$
\binom{n}{k}p^k(1-p)^{n-k}=\mathbb P(S_n=k)=\mathbb P\left(\frac{S_n-\mu_n}{\sigma_n}=x\right),
$$
where $\mu_n$ and $\sigma_n$ are the mean and standard deviation of $X_n$, respectively. By the CLT,
$$
\frac{S_n-\mu_n}{\sigma_n}\to Z\quad\text{ as }\quad n\to\infty,
$$
where $Z$ is a standard normal random variable. Thus for all $\epsilon >0$,
$$
\lim_{n\to\infty}\mathbb P\left(x-\epsilon\leq\frac{S_n-\mu_n}{\sigma_n}\leq x+\epsilon\right)=\mathbb P(x-\epsilon\leq Z\leq x+\epsilon).
$$
Writing out the meaning of the left and right sides explicitly, this says that
$$
\lim_{n\to\infty}\sum_{k=k^-}^{k^+} \binom{n}{k}p^k(1-p)^{n-k}=\frac{1}{\sqrt{2\pi}}\int_{x-\epsilon}^{x+\epsilon} e^{-t^2/2}\ dt,
$$
where ${k^{-}=\Bigl\lfloor np+(x-\epsilon)\sqrt{np(1-p)}\Bigr\rfloor}$ and ${k^{+}=\Bigl\lceil np+(x+\epsilon)\sqrt{np(1-p)}\Bigr\rceil}$.
