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

Using the following inequality 
\begin{equation}
\frac{n}{\sum_{k=1}^{n} \frac{1}{1-\frac{1}{2k}} } \leq \sqrt[n]{(1-\frac{1}{2})(1-\frac{1}{4})...(1-\frac{1}{2n})} \leq \frac{(1-\frac{1}{2})+(1-\frac{1}{4})+...,(1-\frac{1}{2n})}{n} = 1-\frac{1}{2n}\sum_{k=1}^{n}\frac{1}{k}
\end{equation}
and another results
\begin{equation}
\sum_{k=1}^{n}\frac{1}{k} = \ln(n) + \gamma ,n \rightarrow \infty,
\end{equation}
where $\gamma $ is Euler constant. 
and 
\begin{equation}
\sum_{k=1}^{n} \frac{1}{1-\frac{1}{2k}}=\sum_{k=1}^{n}\frac{2k-1 +1}{2k-1} = n + \sum_{k=1}^{n}\frac{1}{2k-1}\leq n + \sum_{k=1}^{n}\frac{1}{k} 
\end{equation}
Hence we have 
\begin{equation}
\frac{n}{n+\ln(n) + \gamma} \leq \sqrt[n]{(1-\frac{1}{2})(1-\frac{1}{4})...(1-\frac{1}{2n})} \leq 1 - \frac{\ln(n) + \gamma}{2n} 
\end{equation}
