asymptotics of central binomial coefficient

By making use of the expression

$$4^{-n}\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=\prod _{m=1}^n\frac{2m-1}{2m},$$

we may obtain estimates of the central binomialcoefficient $\left(\genfrac{}{}{0pt}{}{2n}{n}\right)$ for large values of $n$.We begin by making some elementary algebraicmanipulations of this product:

$$4^{-n}(2n+1)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=\frac{(2n+1)\prod _{m=1}^n(2m-1)}{{\prod }_{m=1}^{n}2m}=\frac{\prod _{m=1}^n(2m+1)}{{\prod }_{m=1}^{n}2m}=\prod _{m=1}^n\frac{2m+1}{2m}$$

Then we multiply this by our previous expression, factor-by-factor:

$${16}^{-n}(2n+1){\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^2=\prod _{m=1}^n\frac{(2m+1)(2m-1)}{{(2m)}^2}.$$

With a bit of rearrangement and manipulation, this givesus the following formula for the central binomial coefficient,

$$\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=\frac{4^n}{\sqrt{2n+1}}\sqrt{\prod _{m=1}^n\left(1-\frac{1}{4m^2}\right)},$$

which we shall examine to obtain estimates of the centralbinomial coefficient for large $n$.

To make use of this formula, we make two key observationsabout the product which appears on the right-hand side.First, since each of the terms in the product lies between$0$ and $1$, the product is an decreasing function of $n$,Thus, we have

$$4^{-a}\sqrt{2a+1}\left(\genfrac{}{}{0pt}{}{2a}{a}\right)>4^{-b}\sqrt{2b+1}\left(\genfrac{}{}{0pt}{}{2b}{b}\right)$$

when . Secondly, the product converges in the limit$n\to \mathrm{\infty }$. In fact, it turns out to be Wallis’ productfor $\pi$, so we have

$$\underset{n\to \mathrm{\infty }}{lim}{4}^{-n}\sqrt{2n+1}\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=\sqrt{\frac{2}{\pi }}.$$

This limit may be reread as an approximate formula when$n$ is large:

$$\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\approx \sqrt{\frac{2}{\pi }}\frac{4^n}{\sqrt{2n+1}}$$

As an example, let us consider the case $n=10$. Theexact answer is $\left(\genfrac{}{}{0pt}{}{20}{10}\right)=184756$. The approximateanswer is $(4^{10}\sqrt{2}/\sqrt{21\pi })=182570.38\mathrm{\dots }$,which agrees with the exact answer to a percent. Alsonote that the estimate is smaller than the exact answer —this is a general feature which is due to the observationmade above that the product is an decreasing function of $n$.Moreover, this observation also implies that the percentageerror decreases as $n$ increases; in particular, the approximationis good within a percent when $n>10$.

It is worth noting that same result can be obtained from Stirling’sformula. In fact, one can deduce Stirling’s formula by a similarline of reasoning.

With a little more work, wee can improve our approximation.We begin by considering the product

$$\prod _{m=1}^n\frac{64m^2-9}{64m^2-25}.$$

On the one hand, we can evaluate this product by factoringthe numerator and denominator and cancelling terms:

${\displaystyle \prod _{m=1}^n}{\displaystyle \frac{64m^2-9}{64m^2-25}}$$={\displaystyle \prod _{m=1}^n}{\displaystyle \frac{(8m-3)(8m+3)}{(8m-5)(8m+5)}}$

$={\displaystyle \frac{5\cdot 11}{3\cdot 13}}\cdot {\displaystyle \frac{13\cdot 19}{11\cdot 21}}\cdot {\displaystyle \frac{21\cdot 27}{19\cdot 29}}\mathrm{\cdots }\mathrm{\cdots }{\displaystyle \frac{(8n-11)(8n+5)}{(8n-13)(8n-3)}}\cdot {\displaystyle \frac{(8n-3)(8n+3)}{(nm-5)(8n+5)}}$

$={\displaystyle \frac{5}{3}}\cdot {\displaystyle \frac{8n+3}{8n+5}}$

From this expression, it follows that the product convergesto $5/3$ as $n\to \mathrm{\infty }$.On the other hand, we can multiply this product by the productwe obtained earlier termwise:

$\left({\displaystyle \prod _{m=1}^n}{\displaystyle \frac{64m^2-9}{64m^2-25}}\right)\left({\displaystyle \prod _{m=1}^n}\left(1-{\displaystyle \frac{1}{4m^2}}\right)\right)$$={\displaystyle \prod _{m=1}^n}{\displaystyle \frac{64m^2-9}{64m^2-25}}\cdot {\displaystyle \frac{4m^2-1}{4m^2}}$

$={\displaystyle \prod _{m=1}^n}{\displaystyle \frac{256m^4-100m^2+9}{256m^4-100m^2}}$

$={\displaystyle \prod _{m=1}^n}\left(1+{\displaystyle \frac{9}{256m^4-100m^2}}\right)$

By taking limits on both sides, we conclude that

$$\prod _{m=1}^{\mathrm{\infty }}\left(1+\frac{9}{256m^4-100m^2}\right)=\frac{10}{3\pi }.$$

Embracing our earlier formula for the central binomialcoefficient with both hands, we obtain

$${16}^{-n}\cdot \frac{80n^2+70n+15}{24n+15}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^2=\prod _{m=1}^n\left(1+\frac{9}{256m^4-100m^2}\right).$$

Juggling terms from one hand to the other, we obtaina new formula for the central binomial coefficient:

$$\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=4^n\sqrt{\frac{24n+15}{80n^2+70n+15}}\sqrt{\prod _{m=1}^n\left(1+\frac{9}{256m^4-100m^2}\right)}$$

Despite the increase in complexity, this formula isactually an improvement over the old formula. The reasonfor this is that the product converges more rapidlybecause the polynomial in the denominator is nowof the fourth order.

As before, we may take the limit $n\to \mathrm{\infty }$:

$$\underset{n\to \mathrm{\infty }}{lim}{4}^{-n}\sqrt{\frac{16n^2+14n+3}{8n+5}}\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=\sqrt{\frac{2}{\pi }}$$

This gives us a new, improved asymptotic formula:

$$\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\approx \sqrt{\frac{2}{\pi }}{\mathrm{\hspace{0.17em}4}}^n\sqrt{\frac{8n+5}{16n^2+14n+3}}$$

To see just how good this formula is, let us revisitthe case $n=10$. Now, we get the approximation$(4^{10}\sqrt{170/1743\pi })=184756.93\mathrm{\dots }$.Because the terms in the new product are greaterthan unity, the approximation is an overestimate,so we round it down to $184756$, which just sohappens to be the exact answer. The approximationis that good!