# 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!

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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 $1$.

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$(\frac1n(\sum_{k=1}^n1+\frac1{2k-1}))^{-1}\leq (\prod_{k=1}^n(1-\frac1{2k}))^{\frac1n}\leq \frac1n(\sum_{k=1}^n1-\frac1{2k})$

This is the Arithmetic geometric, harmonic mean inequality. Now by squeeze theorem you can get the result.

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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?

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Of course, once one has this idea, there is still non-trivial work to show that we haven't been too loose with our estimates and that the true value really does approach the estimated value as $n \to \infty$. –  Hurkyl Feb 21 '13 at 15:28

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 $1$.

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