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!
 A: If $x_n$ is positive and converges to $x> 0$, then
$$
\sqrt[n]{x_1\cdots x_n}=\exp\left(\frac{\log x_1+\ldots+\log x_n}{n} \right)\longrightarrow \exp(\log x)=x
$$
by Cesaro (http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation) and continuity of $\log$ and $\exp$.
Now take 
$$
x_n=\frac{2n-1}{2n}\longrightarrow 1.
$$
This proves that your sequence 
$$\sqrt[n]{x_1\cdots x_n}$$
converges to $1$.
A: $(\frac1n(\sum_{k=1}^n1+\frac1{2k-1}))^{-1}\leq (\prod_{k=1}^n(1-\frac1{2k}))^{\frac1n}\leq \frac1n(\sum_{k=1}^n1-\frac1{2k})$
This is the Arithmetic geometric, harmonic mean inequality. Now by squeeze theorem you can get the result.
A: 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?
A: 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 $1$.
