Is there a way to approximate the terms of $\frac{\left(2n\right)!}{\left(2^nn!\right)^2}$ for successive $n$ as $n$ becomes large?

Question

I have encountered the ratio of the product of the first n odd numbers to the product of the first n even numbers and want to chart its ultimate convergence to zero. If a white noise signal is passed through a cascade of $n$ linear filters, then this ratio is the factor by which the variance of the signal is reduced by the combined action of those $n$ filters. I am, therefore, interested in the rate at which the expression converges such that I can determine the effectiveness of adding more filters.

Of course, doing so requires very large numbers for the numerator and denominator that exceed computing capacity. Is there a way to approximate the terms of $$\frac{\left(2n\right)!}{\left(2^nn!\right)^2}$$ for successive n as n becomes large?

Answer 1

By Stirling's formula, you have $$\frac{(2n)!}{(2^n n!)^2}=\frac{\sqrt{4\pi n}\Bigl(\cfrac{\not2n}{\mathrm e}\Bigr)^{\!2n}}{\not2^{\not2n}\cdot 2\pi n\Bigl(\cfrac{n}{\mathrm e}\Bigr)^{\!2n}}=\frac 1{\sqrt{\pi n}}.$$

Answer 2

Stirling's approximation gives the following asymptotic for the central binomial coefficient: $$ {2n \choose n} \sim \frac{4^n}{\sqrt{\pi n}}\text{ as }n\rightarrow\infty $$ Therefore, $$ \frac{\left(2n\right)!}{\left(2^nn!\right)^2} = \frac{1}{4^n}{2n \choose n} \sim \frac{1}{\sqrt{\pi n}} $$

Answer 3

Hint

Note that $$ \eqalign{ & S_{\,n} = {{\left( {2n} \right)!} \over {\left( {2^{\,n} n!} \right)^{\,2} }} = {{\prod\limits_{k = 0}^{2n - 1} {2n - k} } \over {\left( {\prod\limits_{k = 0}^{n - 1} 2 \prod\limits_{k = 0}^{n - 1} {n - k} } \right)^{\,2} }} = \cr & = {{\prod\limits_{k = 0}^{n - 1} {2n - 2k} \;\prod\limits_{k = 0}^{n - 1} {2n - 2k - 1} } \over {\prod\limits_{k = 0}^{n - 1} {2n - 2k} \; \left( {\prod\limits_{k = 0}^{n - 1} 2 \prod\limits_{k = 0}^{n - 1} {n - k} } \right)}} = \cr & = {{\prod\limits_{k = 0}^{n - 1} {n - k - 1/2} } \over {\;\prod\limits_{k = 0}^{n - 1} {n - k} }} = \cr & = \prod\limits_{k = 0}^{n - 1} {1 - {1 \over {2\left( {n - k} \right)}}} = \prod\limits_{k = 1}^n {1 - {1 \over {2k}}} \cr} $$

Then pass to $\ln S_n$ and to Riemann sum .

Also, restarting from the above we get $$ \eqalign{ & S_{\,n} = {{\prod\limits_{k = 0}^{n - 1} {n - k - 1/2} } \over {\;\prod\limits_{k = 0}^{n - 1} {n - k} }} = {{\prod\limits_{k = 0}^{n - 1} {1/2 + k} } \over {\;\prod\limits_{k = 0}^{n - 1} {1 + k} }} = \cr & = {{\left( {1/2} \right)^{\,\overline {\,n\,} } } \over {1^{\,\overline {\,n\,} } }} = {{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {1/2} \right)}} {{\Gamma \left( 1 \right)} \over {\Gamma \left( {n + 1} \right)}} = {{\left( {1/2} \right)^{\,\overline {\,1/2\,} } } \over {\left( {n + 1/2} \right)^{\,\overline {\,1/2\,} } }} = \cr & = \left( \matrix{ n - 1/2 \cr n \cr} \right) = \left( { - 1} \right)^{\,n} \left( \matrix{ - 1/2 \cr n \cr} \right) \cr} $$ so that $$ \sum\limits_{0\, \le \,n} {S_{\,n} \,x^{\,n} } = {1 \over {\sqrt {1 - x} }} $$ and there is plenty of hints for analyzing the asymptotics at different degree of approximation.

Answer 4

From Wikipedia, we have:

$$\frac{(2n)!}{(2^nn!)^2}=\frac{1}{4^n}\binom{2n}n=\frac1{\sqrt{\pi n}}\left(1-\frac{c_n}{n}\right) $$ where $\frac19<c_n<\frac18.$

Answer 5

In this answer, equation $(10)$ says $$ \frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}} $$ which gives $$ \frac1{\sqrt{\pi\!\left(n+\frac13\right)}}\le\frac{(2n)!}{\left(2^nn!\right)^2}\le\frac1{\sqrt{\pi\!\left(n+\frac14\right)}} $$

Answer 6

Again, without Stirling $$ \frac{2n!}{(2^n n!)^2} = e^{\log (2n)! - 2 \log 2^n n!} \approx e^{2n \log 2n - 2n +1 - 2n \log 2 - 2n \log n+2n -2} =e^{-1} $$

Answer 7

You already received so many good answers that the only thing I could do is to provide a quite good approximation.

$$\frac{(2n)!}{(2^n n!)^2}=\frac 1{\sqrt{\pi n}} \frac {1-\frac{1}{16 n}+\frac{11}{256 n^2} } {1+\frac{1}{16 n}+\frac{11}{256 n^2} }$$ which shows an absolute relative error smaller than $0.001$% as soon as $n \geq 3$ and smaller than $0.0001$% as soon as $n \geq 5$.

Also, in the same spirit as @robjohn, $$\frac{(2n)!}{\left(2^nn!\right)^2}\sim \frac1{\sqrt{\pi\!\left(n+\frac14+\frac 1{32n}-\frac1 {128n^2}-\frac{5}{2048 n^3}+\frac{23}{8192 n^4}\right)}}$$ could provide tighter bounds. It shows an absolute relative error smaller than $0.001$% as soon as $n \gt 1$ and smaller than $0.0001$% as soon as $n \gt 2$.