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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.

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    $\begingroup$ Since $(2n)!!=n!·2^n$ and $(2n+1)!!=\dfrac{(2n+1)!}{(2n)!!}=\dfrac{(2n+1)!}{n!·2^n}$, it reduces to find better bounds for $m!$. $\endgroup$ – Saad Jun 26 '18 at 12:47
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    $\begingroup$ How about using bounds for the Gamma function or the ordinary factorial with the relations \begin{align*} (2k)!! &= 2^k k!\\ (2k+1)!! &= (2k+1)!/(2^k k!) = \frac{2^{k+1}}{\sqrt{\pi}}\,\Gamma\left(k+\tfrac{3}{2}\right) \end{align*} $\endgroup$ – gammatester Jun 26 '18 at 12:49
  • $\begingroup$ @AlexFrancisco thank you, you are right, I have to get better bounds for the factorial. $\endgroup$ – Mauricio Fernández Jun 26 '18 at 13:12
  • $\begingroup$ @gammatester hmmmm, .... that would be intersting, specially using dlmf.nist.gov/5.6#E8 (last equation) $\endgroup$ – Mauricio Fernández Jun 26 '18 at 13:15
  • $\begingroup$ Any thoughts on my answer, Mauricio? $\endgroup$ – Gerry Myerson Jun 28 '18 at 3:17
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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.

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    $\begingroup$ Hi Gerry, I took a look at Veclav's asymptotic. I updated my question, you may want to have a look. Thanks for the reference. It actually gave me an idea for another problem :D $\endgroup$ – Mauricio Fernández Jun 28 '18 at 10:38

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