Convergence of sum with binomial coefficient I found this exercise in a book and I don't know how to start:

With the help of probabilistic methods, prove that
$$
\lim_{n\rightarrow\infty} \frac{1}{4^n} \sum_{k=0}^n  \binom{2n}{k} = 1/2
$$

One cannot use the binomial theorem because the sum just goes to $n$ and not to $2n$ and besides that, am I supposed to use one of the limit theorems in stochastics? If so, which one and how can I approach this exercise.
 A: Let $(\xi_j)_{j\geqslant 1}$ be an i.i.d. sequence of random variables such that $\mathbb P(\xi_1=1)=\mathbb P(\xi_1=0)=1/2$. Then $\sum_{j=1}^{2n}\xi_j$ has a binomial distribution with parameters $(2n,1/2)$ hence
$$
\frac{1}{4^n} \sum_{k=0}^n  \binom{2n}{k}=\mathbb P\left(\sum_{j=1}^{2n}\xi_j\leqslant n\right).
$$
The central limit theorem will give the answer.
A: We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\binom{2n}{k}}&=\frac{1}{2}\sum_{k=0}^{n-1}\binom{2n}{k}+\binom{2n}{n}+\frac{1}{2}\sum_{k=n+1}^{2n}\binom{2n}{k}\tag{1}\\
&=\frac{1}{2}\sum_{k=0}^{2n}\binom{2n}{k}+\frac{1}{2}\binom{2n}{n}\\
&\,\,\color{blue}{=\frac{1}{2}2^{2n}+\frac{1}{2}\binom{2n}{n}}\tag{2}
\end{align*}
In (1) we use the symmetry $\binom{2n}{k}=\binom{2n}{2n-k}$. We use Stirling's approximation formula
\begin{align*}
n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}
\end{align*}
and get
\begin{align*}
\color{blue}{\binom{2n}{n}}=\frac{(2n)!}{n!n!}
&\sim\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}\left(\frac{e}{n}\right)^{2n}\frac{1}{2\pi n}\\
&\,\,\color{blue}{\sim 4^n\frac{1}{\sqrt{\pi n}}}\tag{3}
\end{align*}

We conclude from (2) and (3)
\begin{align*}
\color{blue}{\frac{1}{4^n}}\color{blue}{\sum_{k=0}^n\binom{2n}{k}}
&=\frac{1}{4^n}\left(\frac{1}{2}2^{2n}+\frac{1}{2}\binom{2n}{n}\right)\\
&=\frac{1}{2}+\frac{1}{2}\,\frac{1}{4^n}\binom{2n}{n}\\
&\sim\frac{1}{2}+\frac{1}{2}\frac{1}{\sqrt{\pi n}}\color{blue}{\underset{n\to\infty }{\longrightarrow} \frac{1}{2}}
\end{align*}
and the claim follows.

