# Series Expansion at $n=\infty$ for $\frac{(2n-1)!!}{(2n)!!}$

Looking at this question, I asked Wolfram and got a Series Expansion at $n=\infty$ for $\displaystyle \frac{(2n-1)!!}{(2n)!!}$ like $\displaystyle \left({n^{-1/2}} -\frac{n^{-3/2}}{8}+\frac{n^{-5/2}}{128}-\frac{5n^{-7/2}}{1024}+\frac{21n^{-9/2}}{32768} -\frac{399n^{-11/2}}{262144}+O(n^{-13/2})\right) \pi^{-1/2}$.

Can anybody explain this? Where do these rational coefficients come from?

EDIT2: They don't seem to follow a straight forward pattern:

$\displaystyle \frac{1}{1},-\frac{1}{2^3},\frac{1}{4\times 2^5},-\frac{5}{2^3\times2^7},\frac{21}{2^6\times 2^9},-\frac{399}{2^7\times 2^{11}},\dots$. Is there a closed formula for them?

EDIT: Since this is dealing with integer $n$, I removed the $\cos(2n\pi)$ part, in the exponent of $2/\pi$. And further, wouldn't this give a another bound on the linked question, like $\sqrt{\frac{1}{\pi n} }$?

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Even though the identity $$\frac{(2n-1)!!}{(2n)!!}=\frac{(2n)!}{4^n (n!)^2}$$ holds for positive integers $n$, both sides are interpolated in different ways in Mathematica. Refer to the equation (5) in this site to see how Mathematica understands $x!!$. I presume what you want to know is the series expansion of the function $$\frac{(2n)!!}{4^n (n!)^2}=\frac{\Gamma(2n+1)}{4^n \Gamma(n+1)^2}$$ near $n=\infty$. In this case, we have $$\exp\left[-\frac{1}{2}\log (\pi n)-\frac{1}{8n}+\frac{1}{192n^3}+O\left(\frac{1}{n^5}\right)\right].$$ –  sos440 Mar 13 '12 at 19:45
@sos440 Could you relate your expression with mine (in an answer)? –  draks ... Mar 15 '12 at 11:49

Maple says the definition for double factorial $n!!$ is: $$\mathrm{doublefactorial}(n)=2^{n/2} (2/\pi)^{1/4-1/4 \cos(\pi n)} (n/2)!,$$ and it looks like Mathematica is using this as well.
So, if $n$ is an even integer, $$n!! = 2^{n/2} \biggl(\frac{n}{2}\biggr)!$$ and if $n$ is an odd integer, $$n!! = 2^{(n+1)/2} \sqrt{\frac{1}{\pi}} \biggl(\frac{n}{2}\biggr)!$$ and of course factorial of non-integer is done in terms of the Gamma function. Now, divide and do asymptotics according to Stirling's formula, to get \begin{align} &\frac{(2n-1)!!}{(2n)!!} = \frac{1}{\sqrt{\pi}\;n!} \Bigl(n - \frac{1}{2}\Bigr)! \\ &\qquad= \frac{1}{\sqrt{\pi} \sqrt{n}} - \frac{1}{8 \sqrt{\pi} n^{3/2}} + \frac{1}{128 \sqrt{\pi} n^{5/2}} + \frac{5}{1024 \sqrt{\pi} n^{7/2}} - \frac{21}{32768 \sqrt{\pi} n^{9/2}} + O \Biggl(n^{-11/2}\Biggr) \end{align} There is no simple explanation for these coefficients. The coefficients in Stirling's formula involve Bernoulli numbers. And this is the quotient of two such asymptotic series, done by long division.
If $n$ is an integer, the two interpolation methods agree. That is when $\cos(2 n \pi) = 1$. –  GEdgar Mar 14 '12 at 0:22
Ah, integer $n$ is fine. Thanks. –  draks ... Mar 14 '12 at 7:59
So I get $\displaystyle 2^{1/2} (2/\pi)^{-1/4[ \cos(\pi 2n)-\cos(\pi (2n-1))]} {n!}/{(n-1/2)!}$, when I apply the definition? Some parts of the WA given formula are already visible, but where does my series come from? –  draks ... Mar 14 '12 at 11:45