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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
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1 Answer

up vote 3 down vote accepted

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.

added

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.

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And what does this means for my series expansion? Is it wrong? I also don't see how your answer relates to @sos440's comment. –  draks ... Mar 13 '12 at 22:29
    
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
    
If you can show, how to get to that series, your answer would be "acceptable". Or is this just a tedious work without any nice things to find on the way...? –  draks ... Mar 15 '12 at 11:49
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