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What is the asymptotic behavior of the binomial coefficients near the center? Specifically, $$ {2n \choose n - n \epsilon_n} $$ for some $\epsilon_n \ll 1$. In particular, I am interested in an approximation of the binomial coefficient for the case when $\epsilon_n = 1 / \log n$.

Wikipedia (https://en.wikipedia.org/wiki/Binomial_coefficient) writes an approximation that is valid for $\epsilon_n = o(n^{-1/3})$, but how about larger values of $\epsilon_n$ such as $1 / \log n$?

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  • $\begingroup$ Did you try to apply Stirling formula? This is the first idea that should come to mind... $\endgroup$ – Did Aug 28 '16 at 13:28
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Considering $$A_n={2n \choose n - n \epsilon}=\frac{(2n)!}{(n-n\epsilon)! \,(n+n\epsilon)!}$$ replace the factorials by the corresponding Gamma functions. Taking logarithms and expanding as Taylor series around $\epsilon=0$, one could obtain $$\log(A_n)=(\log ((2 n)!)-2 \log (n!))-n^2 \epsilon ^2 \psi ^{(1)}(n+1)+O\left(\epsilon ^4\right)$$ where appears the first derivative of the digamma function.

Using $\epsilon=\frac 1 {\log(n)}$ and plotting, it seems to be a quite good approximation as show in the table below $$\left( \begin{array}{ccc} n & \log(A_n) & \text{approx} \\ 5 & 3.67380 & 3.77940 \\ 10 & 10.2770 & 10.3318 \\ 15 & 16.8366 & 16.8810 \\ 20 & 23.4355 & 23.4756 \\ 25 & 30.0673 & 30.1053 \\ 30 & 36.7243 & 36.7611 \\ 35 & 43.4009 & 43.4370 \\ 40 & 50.0931 & 50.1288 \\ 45 & 56.7981 & 56.8336 \\ 50 & 63.5137 & 63.5492 \end{array} \right)$$

For sure, we could make better increasing the order of the expansion to give $$\log(A_n)=(\log ((2 n)!)-2 \log (n!))-n^2 \epsilon ^2 \psi ^{(1)}(n+1)-\frac{1}{12} n^4 \epsilon ^4 \psi ^{(3)}(n+1)+O\left(\epsilon ^6\right)$$ The next table reproduces some results for this secod approximation. $$\left( \begin{array}{ccc} n & \log(A_n) & \text{approx} \\ 5 & 3.67380 & 3.68759 \\ 10 & 10.2770 & 10.2809 \\ 15 & 16.8366 & 16.8389 \\ 20 & 23.4355 & 23.4372 \\ 25 & 30.0673 & 30.0688 \\ 30 & 36.7243 & 36.7256 \\ 35 & 43.4009 & 43.4020 \\ 40 & 50.0931 & 50.0941 \\ 45 & 56.7981 & 56.7990 \\ 50 & 63.5137 & 63.5146 \end{array} \right)$$

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