Proof of inequality with ratio of odds to evens. A professor gave this to the class as a challenge. It may or may not have to do with calculus, but I'm running out of ideas. 
Prove that for any $n > 1$
$$ \frac{1}{2\sqrt{n}} < \frac{1}{2} \cdot \frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2n -1}{2n} < \frac{1}{\sqrt{2n}} $$

When I first looked at this I though it was simply something I could prove with the mean value theorem. Where 
$$ \frac{2n -1}{2n} - \frac{1}{\sqrt{2n}} < 0 $$
for any $x > 1$. 
I made some progress on that but then I realized that $ \frac{2n -1}{2n} $ is not a formula for the middle term that I could treat as a function, but rather the product of the terms up to $\frac{2n -1}{2n}$, so you can imagine my disappointment. 
With this new realization and as I had already seen with the previous approach, I focused on
$$ \frac{2n -1}{2n} = 1 - \frac{1}{2n}. $$ Simple enough.
Which makes the middle expression 
$$ \left(1 - \frac{1}{2(1)} \right)\cdot \left(1 - \frac{1}{2(2)} \right)\cdot \left(1 - \frac{1}{2(3)} \right)\cdots $$
But I'm not sure what to make of this product. It reminded me of the golden ratio definition and maybe of the divisor function, but I might be readding too much into it. 
I also see that 
$$ 2n - 1 < 2n $$
For $n > 1$. So the ratio 
$$ \frac{2n -1}{2n} < 1 $$ 
and likely the product 
$$ \left(1 - \frac{1}{2(1)} \right)\cdot \left(1 - \frac{1}{2(2)} \right)\cdot \left(1 - \frac{1}{2(3)} \right)\cdots  < 1 $$
But I'm not sure where to go from there. 
I'd appreciate any help.
 A: Let $\displaystyle \frac1A = \frac{1}{2} \cdot \frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2n -1}{2n}$.  Then we have 
$$\displaystyle \frac{1}{2\sqrt{n}} < \frac1A < \frac{1}{\sqrt{2n}} \iff 2\sqrt{n} > A > \sqrt{2n} \iff 4n > A^2 > 2n$$
For the lower bound, we have to show
$$\frac{2^2}{1} \cdot \frac{4^2}{3^2}\cdot \frac{6^2}{5^2} \cdots \frac{(2n)^2}{(2n-1)^2} > 2n$$
$$\iff \frac{2^2}{1 \cdot 3} \cdot \frac{4^2}{3\cdot 5}\cdot \frac{6^2}{5\cdot 7} \cdots \frac{(2n)^2}{(2n-1)\cdot(2n+1)} > \frac{2n}{2n+1}$$
$$\iff \left(\prod_{k=1}^{n-1} {\frac{(2k)^2}{(2k)^2-1}}\right)\frac{2n}{2n-1} > 1$$
As we can note every factor on the LHS is $> 1$, the inequality holds.
Similarly one can work for the upper bound to get the equivalent:
$$ 2n > \frac{2\cdot 4}{3^2} \cdot \frac{4\cdot 6}{5^2}\cdot \frac{6\cdot 8}{7^2} \cdots \frac{(2n-2)(2n)}{(2n-1)^2}(2n) $$
$$\iff 1 > \prod_{k=1}^{n-1} \frac{2k (2k+2)}{(2k+1)^2}$$
which is again evident as all factors in the RHS are $< 1$.
A: hint: if you multiply both numerator and denominator by the denominator then you see that the fraction is equal to:
$$
\frac{(2n)!}{2^{2n}(n!)^2}
$$
