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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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    @sos440 Could you relate your expression with mine (in an answer)?2012-03-15

1 Answers 1

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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.

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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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    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...?2012-03-15