Evaluate $f(n)=\frac{(n-2)!!}{(n-3)!!}$, for $n \geq 3$ by Big O notation 
Is there a better way than mine to evaluate this function
$$f(n)=\frac{(n-2)!!}{(n-3)!!},\quad \text{for}\quad n \geq 3$$
by Big O notation?

I got result below, but not so accurate. I know that:
$$n!!=\left\{ \begin{array}{ll}
       2^{k}k! &,n=2k\\
       \frac{(2k-1)!}{2^{k-1}(k-1)!} & , n=2k-1\\
       \end{array} \right.$$
So, I put this in $f(x)$ and got (for $x=\frac{n-2}{2}$, because for $n=2k$ function is always bigger in this case):
$$f(x)=\frac{2^{2x-1}x!(x-1)!}{(2x-1)!}$$
When I use asymptotic equation for $x$!:
$$x! = \sqrt{2\pi x}\bigg(\frac{x}{e}\bigg)^x\bigg(1+\frac{1}{12x}+\frac{1}{288n}+O(x^{-3})\bigg)$$
And notice that $e\gt2$ (I must somehow evalute $2^{2x-1}$ by Big O):
$$e^{2x-1} = 1 + 2x-1 + \frac{(2x-1)^2}{2!}+ O(x^{-3})$$
So now I connect everything, but for $n = 6$ i got error $0,3$ and it's too much
(because of my "fancy" evaluation $e\gt 2$). When I paste this into Wolfram it looks like a $O(\log)$ or $O(\sqrt{n})$ but I can't figure this out how to get it.
 A: If we consider $$f(x)=\frac{2^{2x-1}\,x!\,(x-1)!}{(2x-1)!}$$ taking logarithms, using Stirling approximation and continuing with Taylor expansions, we have
$$\log[f(x)]=\frac{1}{2} \log (\pi  x)+\frac{1}{8 x}-\frac{1}{192 x^3}+\frac{1}{640
   x^5}+O\left(\frac{1}{x^7}\right)$$
$$f(x)=\sqrt{\pi x}\Bigg[1+\frac{1}{8 x}+\frac{1}{128 x^2}-\frac{5}{1024 x^3}-\frac{21}{32768
   x^4}+O\left(\frac{1}{x^5}\right) \Bigg]$$
For $x=2$, the absolute error is $1.09\times 10^{-4}$ and it decreases very fast (for $n=20$, it is $3.79\times 10^{-9}$).
A: You should analyze the even and odd cases separately since they have different asymptotics.
Also, there is no need to approximate $2^{2n}$ since it cancels out.
If $n = 2k$ we have
$$\frac{(2k)!!}{(2k-1)!!}=\frac{(2^k k!)^2}{(2k)!} = \frac{4^k\left(\sqrt{2\pi k}\left(\frac{k}{e}\right)^k\left(1+\frac{1}{12k}+\frac{1}{288k^2}+O\left(\frac{1}{k^3}\right)\right)\right)^2}{\sqrt{2\pi 2k}\left(\frac{2k}{e}\right)^{2k}\left(1+\frac{1}{12\cdot 2k}+\frac{1}{288\cdot 4k^2}+O\left(\frac{1}{k^3}\right)\right)}=\sqrt{\pi k}\frac{\left(1+\frac{1}{12k}+\frac{1}{288k^2}+O\left(\frac{1}{k^3}\right)\right)^2}{1+\frac{1}{12\cdot 2k}+\frac{1}{288\cdot 4k^2}+O\left(\frac{1}{k^3}\right)}=\sqrt{\pi k}\left(1+\frac{1}{8k}+\frac{1}{128k^2}+O\left(\frac{1}{k^3}\right)\right)$$
A similar calculation can be done in the case of $n=2k+1$.
