What is the value of $\lim_{n\to \infty} \sum_{j=n}^{4n} {4n \choose j}\bigl(\frac{1}{4}\bigr)^{j} \bigl(\frac{3}{4}\bigr)^{4n-j}$? Find the value of the limit
$$\lim_{n\to \infty} \sum_{j=n}^{4n} {4n \choose j}\left(\dfrac{1}{4}\right)^{j} \left(\dfrac{3}{4}\right)^{4n-j}$$
First we take the limit of the summation from $0$ to $\infty$ then subtract the sum from $0$ to $n-1$. The value of the first sum is $1$. I can't determine the second sum with limit. How we find it?
 A: We can regard such a sum as the probability that a binomial variable $B$, given by the sum of $4n$ independent Bernoulli variables $X_i$ with $p=\frac{1}{4}$, lies in the interval $[n,4n]$. Since the expected value of $B$ is $n$, the strong law of large numbers gives that the value of the limit is just $\color{red}{\frac{1}{2}}$.
The second term of the asymptotics obviously depends on the skewness of $X_i$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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See this question:
\begin{align}&\color{#66f}{\large%
\lim_{n\ \to\ \infty}\ \sum_{j\ =\ n}^{4n}{4n \choose j}
\pars{1 \over 4}^{j}\pars{3 \over 4}^{4n-j}}
=\lim_{n\ \to\ \infty}\
{\,{\rm B}\pars{1/4,n,3n + 1} \over \,{\rm B}\pars{n,3n + 1}}=\color{#66f}{\large\half}
\end{align}

$\ds{\,{\rm B}}$'s are Beta Functions with
  $\ds{\,{\rm B}\pars{x,y}=\,{\rm B}\pars{1,x,y}}$.

Note que
\begin{align}
\,{\rm B}\pars{p,n,3n + 1}&=\int_{0}^{p}t^{n - 1}\,\pars{1 - t}^{3n}\,\dd t
\\[5mm]&\sim\pars{1 \over 4}^{n - 1}\pars{3 \over 4}^{3n}\root{2}\sigma_{n}
\int_{- 1/\pars{4\root{2}\sigma_{n}}}^{\pars{p - 1/4}/\pars{\root{2}\sigma_{n}}}
\expo{-t^{2}}\,\dd t\,,\quad\sigma_{n}\sim{\root{3} \over 8}\,n^{-1/2}
\\[5mm]&\mbox{when}\quad n \gg 1
\end{align}

The above limit is equivalent to
  $$
{\ds{\int_{-\infty}^{0}\expo{-t^{2}}\,\dd t}\over
 \ds{\int_{-\infty}^{\infty}\expo{-t^{2}}\,\dd t}} = \half
$$

