By making use of the expression $$ 4^{-n} {2n \choose n} = \prod_{m=1}^n {2m - 1 \over 2m} , $$ we may obtain estimates of the central binomial coefficient ${2n \choose n}$ for large values of $n$ . We begin by making some elementary algebraic manipulations of this product: $$ 4^{-n} (2n+1) {2n \choose n} = {(2n+1) \prod_{m=1}^n (2m-1) \over \prod_{m=1}^n 2m} = {\prod_{m=1}^n (2m+1) \over \prod_{m=1}^n 2m} = \prod_{m=1}^n {2m+1 \over 2m} $$ Then we multiply this by our previous expression, factor-by-factor: $$ 16^{-n} (2n+1) {2n \choose n}^2 = \prod_{m=1}^n {(2m+1) (2m-1) \over (2m)^2} . $$ With a bit of rearrangement and manipulation, this gives us the following formula for the central binomial coefficient, $$ {2n \choose n} = {4^n \over \sqrt{2n+1}} \sqrt{ \prod_{m=1}^n \left( 1 - {1 \over 4 m^2} \right)} , $$ which we shall examine to obtain estimates of the central binomial coefficient for large $n$ .

To make use of this formula, we make two key observations about 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 $$ 4^{-a} \sqrt{2a+1} {2a \choose a} > 4^{-b} \sqrt{2b+1} {2b \choose b} $$ when $a < b$ . Secondly, the product converges in the limit $n \to \infty$ . In fact, it turns out to be Wallis' product for $\pi$ , so we have $$ \lim_{n \to \infty} 4^{-n} \sqrt{2n+1} {2n \choose n} = \sqrt{2 \over \pi} . $$

This limit may be reread as an approximate formula when $n$ is large: $$ {2n \choose n} \approx \sqrt{2 \over \pi} {4^n \over \sqrt{2n+1}} $$ As an example, let us consider the case $n=10$ . The exact answer is ${20 \choose 10} = 184756$ . The approximate answer is $(4^{10} \sqrt{2}/\sqrt{21 \pi}) = 182570.38\ldots$ , which agrees with the exact answer to a percent. Also note that the estimate is smaller than the exact answer -- this is a general feature which is due to the observation made above that the product is an decreasing function of $n$ . Moreover, this observation also implies that the percentage error decreases as $n$ increases; in particular, the approximation is good within a percent when $n > 10$ .

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

With a little more work, wee can improve our approximation. We begin by considering the product $$ \prod_{m=1}^n {64 m^2 - 9 \over 64 m^2 - 25} . $$ On the one hand, we can evaluate this product by factoring the numerator and denominator and cancelling terms:

As before, we may take the limit $n \to \infty$ : $$ \lim_{n \to \infty} 4^{-n} \, \sqrt{16 n^2 + 14 n + 3 \over 8 n + 5} {2n \choose n} = \sqrt{2 \over \pi} $$ This gives us a new, improved asymptotic formula: $$ {2n \choose n} \approx \sqrt{2 \over \pi} \, 4^n \, \sqrt{8 n + 5 \over 16 n^2 + 14 n + 3} $$

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