How to evaluate $\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$ Im tempted to say that the limit of this sequence is 1 because infinite root of infinite number is close to 1 but maybe Im mising here something? What will be inside the root?
This is the sequence:  
$$\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$$

It is quite easy to show by induction that:
$$\frac{(2n-1)!!}{(2n)!!}\geq\frac{1}{\sqrt{2n}}\tag{1}$$
since the last line is implied by:
$$\frac{2n+1}{2n+2}\geq\sqrt{\frac{2n}{2n+2}}$$
that is equivalent to:
$$(2n+1)^2\geq 2n(2n+2) = (2n+1)^2-1.$$
Using $(1)$ and the trivial bound $\frac{(2n-1)!!}{(2n)!!}\leq 1$, it follows that the limit is $1$ by squeezing.
