How can we prove the Inequality : $ \frac {n!}{ 2^{n-1}((\frac {n-1}{2})!)^2} \leq \sqrt{n}$ How can we prove the following inequality? For every odd positive integer $n$,
$$ \frac {n!}{ 2^{n-1}((\frac {n-1}{2})!)^2} \leq \sqrt{n}$$
Thank You.
 A: In my answer here, I show that 
$${2n\choose n}{1\over 4^n}\leq {1\over\sqrt{\pi n}}.$$
Substitute $(n-1)/2$ for $n$ to get
$${n-1\choose (n-1)/2}{1\over 2^{n-1}}\leq {1\over\sqrt{\pi (n-1)/2}}\leq{1\over\sqrt{n}},$$
the final inequality being valid for  $n\geq 3$, since $\pi (n-1)/2>n$ for such $n$.  
A: Hint:
For $k\geq0$ we need to show $$a_k=\frac{(2k+1)!}{2^{2k}(k!)^2}\leq\sqrt{2k+1}=b_k$$
Now $$a_{k+1}=\frac{(2k+3)!}{2^{2k+2}((k+1)!)^2}= \frac{(2k+3)(2k+2)}{4(k+1)^2}\cdot \frac{(2k+1)!}{2^{2k}(k!)^2}=\frac{(2k+3)(2k+2)}{4(k+1)^2}\cdot a_k$$
So if we knew $a_k\leq b_k$ for some $k$ then could we perhaps continue with induction...?

A nicer proof might be hidden in the in the binomial formula - something like
$$ 2^{2k} = (1+1)^{2k} =\sum_{j=0}^{2k} \frac{(2k)!}{j! (2k-j)!}$$ 
where the mid term is of interest...
A: As Gerry Mentioned, Stirling formula will show it (as this is only an asymptotic behaviour we first needs a lower bound for which we know the equation is true). If you don't want to use such a hammer, you  should try it with induction. 
A: For completeness, here is the proof by induction:
Base Case:  The case $ n = 1$ is easy to check.
Inductive Step:  Assume that $$\frac{k!}{2^{k-1} \left( \frac{k-1}{2} \right)!^2} \leq \sqrt{k}.$$  We aim to show that this implies $$\frac{(k+2)!}{2^{k+1} \left( \frac{k+1}{2} \right)!^2} \leq \sqrt{k+2}.$$
We have
\begin{align}
\frac{(k+2)!}{2^{k+1} \left( \frac{k+1}{2} \right)!^2} &= \frac{(k+2)(k+1)}{4(\frac{k+1}{2})^2} \cdot \frac{k!}{2^{k-1} \left( \frac{k-1}{2} \right)!^2}
\\
&\leq \frac{(k+2)(k+1)}{4(\frac{k+1}{2})^2} \cdot \sqrt{k}
\\
&= \frac{k+2}{k+1} \sqrt{k}
\\
&\leq \sqrt{k+2}
\end{align}
which occurs if and only if $\left( \frac{k+2}{k+1}\right)^2 k \leq k+2$.  I will leave the last inequality for you to check.
