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}  }
$?
 A: 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.
