TITLE: limit of $\sqrt[n]{\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}}$ using geometric mean
QUESTION [8 upvotes]: How to find this question using Geometric Mean?
$$\sqrt[n]{\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}}$$
Thanks!

REPLY [2 votes]: If you group the factors inside the $\sqrt[n]{\cdots}$ properly, you will notice:
$$\sqrt[n]{\frac{1}{2n}} \le \sqrt[n]{\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}} \le 1$$
Since $\lim_{n\to\infty} \sqrt[n]{\frac{1}{2n}} = 1$, your sequence also converges to $1$.