I have been able to prove that this expression, $\frac{\binom {200}{100}}{4^{100}}$ is less than $1$ but I need some fine bounds like if it is less than $\frac{1}{3}$ or between $\frac{1}{3}$ to $\frac{1}{2}$ or less than $\frac{1}{2}$. I calculated the power of $2$ in the numerator and in the denominator and it helped me to get the ratio less than $1$ but I need a strict bound.
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using Stirling's formula $n!\sim\sqrt{2n\pi}\left(\dfrac{n}e\right)^n$ with $n=200 $ and $100$ shows $ \dfrac{200\choose100}{2^{200}}\sim\dfrac1{\sqrt{10}\pi}\approx0.0564$ – J. W. Tanner Aug 25 '19 at 13:47
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Making the problem more general, consider $$b_n=\frac{\binom {2n}{n}}{a^{n}}\implies \log(b_n)=\log \left(\frac{(2 n)! }{a^n \,(n!)^2}\right)=\log((2n)!)-2 \log(n!))-n \log(a)$$
As said in answers and comments, using Stirling approximation gives $$\log(b_n)=n (2 \log (2)-\log (a))-\frac{1}{2} \log \left(\pi{n}\right)-\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$ For the nice case where $a=4$, the first term disappears and we are left with $$\log(b_n)=-\frac{1}{2} \log \left(\pi{n}\right)-\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$ that is to say $$b_n \sim \frac{e^{-\frac{1}{8 n}}}{\sqrt{\pi n } }$$ For $n=100$, $e^{-\frac{1}{800}} \approx 0.998751$ (corresponding to the error given by Taby Mak in his answer).
Claude Leibovici
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From this Math SE post, since $1 - \frac{c_n}{n} ≤ 1$, we have:
$${2n \choose n} < \frac{4^n}{\sqrt{\pi n}}$$
For your case where $n = 100$, this is much greater than the $n > 0.831$ needed to satisfy the inequality.
So we have:
$$\frac{\binom {200}{100}}{4^{100}} < \frac{1}{\sqrt{100 \pi}} = \frac{1}{10 \sqrt \pi}$$
This value is accurate to the original fraction by $0.125 \%$.
Toby Mak
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