I am looking for bounds on the double factorial for even and odd $n \in \mathbb{N}$, defined as
$$
n!! = n \cdot (n-2) \cdot (n-4) \dots
$$
For example, $9!! = 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$ and $8!! = 8 \cdot 6 \cdot 4 \cdot 2$, see Double factorial. It is clear for me that
$$
n! = (n!!)((n-1)!!)
$$
holds, such that if I have some bounds for the factorial
$$
b^-_n \leq n! \leq b^+_n
$$
I can obtain bounds for the double factorial
$$
\frac{b^-_n}{(n-1)!!} \leq n!! \leq \frac{b^+_n}{(n-1)!!}
$$
which can be used recursively, e.g.
$$
\frac{b^-_1 b^-_3 b^-_5}{b^+_2 b^+_4}
\leq
3!!
\frac{b^-_5}{b^+_4}
\leq
\frac{b^-_5}{4!!}
\leq
5!!
\leq
\frac{b^+_5 b^+_3 b^+_1}{b^-_4 b^-_2}
\ .
$$
I have used bounds $b^\pm_n$ based on Stirling's approximation, see Speed of convergence and error estimates, but due to the recursive use, the estimate is too far away for large $n$.
Question: do you know better (explicit) bounds for the double factorial?
Edit (2018-06-28):
In order to provide a clearer view concerning Alex's comment, gammatester comment and and Gerry's answer (see his answer), consider the factorial bounds, see Stirling's approximation
$$
b^\pm_n = c^\pm_n \beta_n
\ , \quad
\beta_n = n^{n+1/2} e^{-n}
\ , \quad
c^-_n = \sqrt{2\pi}
\ , \quad
c^+_n = e
\ .
$$
(It should be remarked that alternative bounds are also given by $c^-_n = \sqrt{2\pi} e^{1/(12n+1)}$ and $c^+_n = \sqrt{2\pi} e^{1/(12n)}$, see Speed of convergence and error estimates). The double factorial of even numbers, as remarked by Alex (see his comment), can be expressed and bounded by the factorial bounds as
$$
(2k)!! = k! \cdot 2^k \Rightarrow 2^k b^-_{k} \leq (2k)!! \leq 2^k b^+_{k} \quad k \in \mathbb{N}
\ .
$$
The asymptotic
$$\alpha_n = n^{(n+1)/2} e^{-n/2}$$
of Vaclav Kotesovec pointed at by Gerry fulfills for even $n=2k, k \in \mathbb{N}$
$$
2^k b^-_{k} = \sqrt{\pi}\alpha_{2k}
$$
such that the asymptotic delivers lower and upper bounds ($2^k b^+_k = \sqrt{2e}\alpha_{2k}$) for even $n=2k$. Interestingly, for odd $n = 2k+1$, the Vaclav's asymptotic seems to fulfill (I just checked for $k=0,1,2,3,\dots,10^4$ in Mathematica)
$$
\sqrt{2} \alpha_{2k+1} \leq (2k+1)!! \leq \sqrt{e} \alpha_{2k+1}
$$
but I am do not know, if this always holds, or something better than $e$ for an upper bound is possible. Alternatively, you can use the bounds based on the relations to the factorial, see Alex's comment,
$$
(2k+1)!!
=
\frac{(2k+1)!}{k! \cdot 2^k}
\geq
\frac{b^-_{2k+1}}{k! \cdot 2^k}
\geq
\frac{b^-_{2k+1}}{b^+_k \cdot 2^k}
\ .
$$
If you are interested in the use of the $\Gamma(x)$ functions, as suggested by gammatester (see his comment), you may want to use the results of the work of Necdet Batir or Necdet Batir (2017). Thanks to all for the suggestions.
Best Answer
The double factorial is tabulated at http://oeis.org/A006882 where there are links to the literature and the asymptotic expression $cn^{(n+1)/2}e^{-n/2}$ where $c=\sqrt{\pi}$ if $n$ is even, $\sqrt2$ if $n$ is odd, attributed to Vaclav Kotesovec.