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How do I approximate ${2n \choose n}$ using Stirling's formula (which approximates ${n!}$ with $\pi$ and $e$?

Siminore
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wolhum jett
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    Do you know how to express $2n \choose n$ in terms of $2n!$ and $n!$? – Scott Lawrence Sep 19 '15 at 15:30
  • do a glimpse at https://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_and_asymptotic_formulas – janmarqz Sep 19 '15 at 15:35
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    Some of the answers here use Stirling: http://math.stackexchange.com/questions/58560/elementary-central-binomial-coefficient-estimates –  Sep 19 '15 at 15:54

3 Answers3

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We can almost avoid Stirling's approximation in providing tight bounds for the central binomial coefficient. The key ingredient is the following identity: $$ \frac{1}{4^n}\binom{2n}{n} = \frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{1}. $$ If we consider the square of the general term of the last product, we get something that behaves like $\left(1-\frac{1}{k}\right)$ for large $k$s, but the product of such terms leads to a telescopic product. That is the motivation behind the following manipulation: $$ \left(\frac{1}{4^n}\binom{2n}{n}\right)^2=\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4n}\prod_{k=1}^{n-1}\left(1-\frac{1}{(2k+1)^2}\right)^{-1}.\tag{2}$$ By the Weierstrass product for the cosine function/ the Wallis product we have: $$\prod_{k=1}^{+\infty}\left(1-\frac{1}{(2k+1)^2}\right)^{-1}=\frac{4}{\pi},\tag{3}$$ hence it follows that: $$ \left(\frac{1}{4^n}\binom{2n}{n}\right)^2=\frac{1}{\pi n}\cdot\prod_{k\geq n}\left(1-\frac{1}{(2k+1)^2}\right)=\frac{1}{\pi n}\cdot\prod_{k\geq n}\left(1+\frac{1}{2k(2k+2)}\right)^{-1}\tag{4} $$ and the last product is clearly $1+O\left(\frac{1}{n}\right)$. By exploiting the inequality $\frac{1}{1+x}\geq e^{-x}$ over $[0,1]$ and a telescopic sum, from $(4)$ we get:

$$ \frac{1}{\sqrt{\pi n}}\geq \frac{1}{4^n}\binom{2n}{n}\geq \frac{1}{\sqrt{\pi n}}\,\exp\left(-\frac{1}{8n}\right).\tag{5}$$

Jack D'Aurizio
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Exactly as one would expect. Since $$\binom{2n}{n}=\frac{(2n)!}{(n!)^2}$$ you can use Stirling's approximation as follows:

$$ (2n)! \operatorname*{\sim}_{n\to\infty} 2\sqrt{\pi n}\frac{(2n)^{2n}}{e^{2n}} $$

$$ (n!)^2 \operatorname*{\sim}_{n\to\infty} 2\pi n\frac{n^{2n}}{e^{2n}} $$

so that

$$ \binom{2n}{n} \operatorname*{\sim}_{n\to\infty} \frac{2\sqrt{\pi n}}{2\pi n}\cdot \frac{e^{2n}}{n^{2n}} \cdot \frac{(2n)^{2n}}{e^{2n}} = \frac{2^{2n}}{\sqrt{\pi n}} $$

Clement C.
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Obviously: $$ \binom{2n}{n}=\frac{(2n)!}{n!n!}$$ I will apply the approximation obtained by H. Robbins in his A Remark on Stirling’s Formula (American Mathematical Monthly 62, 26-29, 1955). Define two functions: $$ f(x)=\sqrt{2\pi}x^{x+\frac{1}{2}}e^{-x}e^{\frac{1}{12x}}$$ $$ g(x)=\sqrt{2\pi}x^{x+\frac{1}{2}}e^{-x}e^{\frac{1}{12x+1}}$$ From the paper I've mentioned we see that $$g(n)<n!<f(n).$$ Thus: $$\frac{g(2n)}{f(n)f(n)}<\binom{2n}{n}=\frac{(2n)!}{n!n!}<\frac{f(2n)}{g(n)g(n)}$$ After substituting with the formulas we see that: $$\frac{\sqrt{2\pi}(2n)^{2n+\frac{1}{2}}e^{-2n}e^{\frac{1}{24n+1}}}{\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}e^{\frac{1}{12n}}\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}e^{\frac{1}{12n}}}<\binom{2n}{n}$$ I think that the simplification and the upper bound you can obtain by yourself.