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:  
$$\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$$

Taking log and Cesaro-Stolz:
$$\eqalign{\log L &=\lim_{n\to\infty}\frac{(\log(1)+\log(3)+\ldots+\log(2n-1))-(\log(2)+\log(4)+\ldots+\log(2n))}n=\cr
&=\lim_{n\to\infty}\frac{\log(2n+1)-\log(2n+2)}1=0.}$$
