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

Your product is itself a geometric mean.
\[ \sqrt[n]{\frac{1\cdot 3\cdot \cdot (2n-1)}{2\cdot 4\cdot \cdot (2n)}}
= \left(\frac{1}{2}\right)^{1/n} \cdot  \left(\frac{3}{4}\right)^{1/n} \dots  \left(\frac{2n-1}{2n}\right)^{1/n}\]
The factor is approaching the same number:
\[ \frac{2n-1}{2n} = 1 - \frac{1}{2n} \to 1\]
If you take geometric mean 1 , 1 and ... 1, what should the answer be?
