Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 4 down vote accepted

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 ( 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 $1$.

share|cite|improve this answer

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

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

share|cite|improve this answer

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?

share|cite|improve this answer
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 $1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.