As part of a some probability problem (probability of gettin $n$ heads and $n$ tails in $2n$ trials with a fair coin) I have derived the formula: $$p_n=\binom{2n}{n}\left(\frac{1}{4}\right)^n$$
I want to evaluate:
$$\lim_{n\to \infty} p_n$$
for $n\in \mathbb N$, using the following bound:
$$e^{\frac{1}{12n+1}}<\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}<e^{\frac{1}{12n}}$$
This is a "special" case of Stirling's approximation from Herbert Robbins that can be found here. I am stuck because I am not sure if my calculation is correct and the result makes no sense to me. When I calculate the limit I get zero. Does this not contradict the law of large numbers? Wouldn't I expect to get a distribution of $50 \%$ heads and $50\%$ tails if I toss a fair coin an infinite amount of times which then should imply $\lim_{n \to \infty} p_n=1$?
What I have tried so far:
$$\begin{equation*}\begin{split}\lim_{n\to \infty} p_n&=\lim_{n\to \infty} \binom{2n}{n} \left(\frac{1}{4}\right)^n \\[10pt] &=\lim_{n \to \infty} \frac{(2n)!}{(n!)^24^n} \end{split}\end{equation*}$$
Rewriting the approximation:
$$ \begin{equation*}\begin{split} &\phantom{\iff} \, \, \, \,e^{\frac{1}{12n+1}}<\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}<e^{\frac{1}{12n}} \\[10pt] & \iff e^{\frac{1}{12n+1}} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n < n!<e^{\frac{1}{12n}} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\end{split}\end{equation*}$$
Replacing $n!$ in $p_n$
$$\begin{equation*}\begin{split} &\phantom{iff}\frac{(2n)!}{\left(e^{\frac{1}{12n+1}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n}<\frac{(2n)!}{(n!)^2 4^n}<\frac{(2n)!}{\left(e^{\frac{1}{12n}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n} \\[10pt] &\iff\underbrace{\frac{1}{\left(e^{\frac{1}{12n+1}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n}}_{:=g(n)}<\underbrace{\frac{1}{(n!)^2 4^n}}_{:=f(n)}<\underbrace{\frac{1}{\left(e^{\frac{1}{12n}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n}}_{:=h(n)}\end{split}\end{equation*}$$
From the squeeze theorem: If $\lim_{n\to \infty} g(n)=\lim_{n \to \infty} h(n)=L$ and $g(n) < f(n) < h(n)$ then $\lim_{n \to \infty} f(n)=L$
$$\begin{equation*}\begin{split}\lim_{n \to \infty} g(n) &= \lim_{n \to \infty} \frac{1}{\left(e^{\frac{1}{12n+1}}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2 4^n} \\[10pt] &=\frac{1}{2 \pi}\lim_{n \to \infty} \frac{1}{e^{\frac{2}{12n+1}} e^{(-2n)} n^{(2n+1)}4^n} \\[10pt] &= \frac{1}{2\pi} \lim_{n \to \infty }\frac{e^{-\frac{24n}{12n+1}}}{n^{2n+1}4^n}=0\end{split}\end{equation*}$$
Analogous $\lim_{n \to \infty} h(n)=0 \implies \lim_{n \to \infty } f(n)=0$. Therefore,
$$\lim_{n \to \infty}p_n=0$$
Best Answer
The correct estimations are as follows: $$ \frac{{(2n)!}}{{n!^2 }} \le \frac{{(2n)^{2n} e^{ - 2n} \sqrt {2\pi 2n} e^{\frac{1}{{24n}}} }}{{\left( {n^n e^{ - n} \sqrt {2\pi n} e^{\frac{1}{{12n + 1}}} } \right)^2 }} = \frac{{4^n }}{{\sqrt {\pi n} }}e^{\frac{1}{{24n}} - \frac{2}{{12n + 1}}} $$ and $$ \frac{{(2n)!}}{{n!^2 }} \ge \frac{{(2n)^{2n} e^{ - 2n} \sqrt {2\pi 2n} e^{\frac{1}{{24n + 1}}} }}{{\left( {n^n e^{ - n} \sqrt {2\pi n} e^{\frac{1}{{12n}}} } \right)^2 }} = \frac{{4^n }}{{\sqrt {\pi n} }}e^{\frac{1}{{24n + 1}} - \frac{1}{{6n}}} . $$ Thus, $$ \frac{1}{{\sqrt {\pi n} }}e^{\frac{1}{{24n + 1}} - \frac{1}{{6n}}} \le p_n \le \frac{1}{{\sqrt {\pi n} }}e^{\frac{1}{{24n}} - \frac{2}{{12n + 1}}} . $$ In particular, $$ p_n \sim \frac{1}{{\sqrt {\pi n} }} $$ as $n\to +\infty$.