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How do you prove the following?

$$\lim_{n\,\to\,\infty}\,\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\ =\ 0$$

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4 Answers 4

Take logarithms: $$\ln\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=\sum_{k=1}^n\ln\frac{2k-1}{2k}=\sum_{k=1}^n\ln\Bigl(1-\frac1{2k}\Bigr)$$ and use $$\ln(1-x)=-x+O(x^2)$$ together with the divergence of the harmonic series.

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We verify easily that for $k\geq 1$, we have $\displaystyle \frac{2k-1}{2k}\leq \sqrt{\frac{k}{k+1}}$. Hence $$\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \prod_{k=1}^n \sqrt{\frac{k}{k+1}}= \frac{1}{\sqrt{n+1}}$$ and we are done.

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Multiply both numerator and denominator by $(2n)!!$, rewrite the denominator as $n! 2^n$ then use Stirling's formula for factorial and the upper bound. What do you get?

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$$\prod_{n=1}^N \frac{2n-1}{2n} = \exp\left\{ \log \left( \prod_{n=1}^N \frac{2n-1}{2n} \right)\right\} = \exp\left\{ \sum_{n=1}^N \log \left(\frac{2n-1}{2n} \right)\right\}$$

$$ = \exp\left\{ \sum_{n=1}^N \log(2n-1) - \log(2n)\right\} = \exp\left\{ \sum_{n=1}^N \log(2n)+\log(1-\frac{1}{2n}) - \log(2n)\right\} = \exp\left\{ \sum_{n=1}^N \log(1-\frac{1}{2n})\right\} $$

For large $n$ the log approximates as $\log(1-\frac{1}{2n}) \approx \frac{-1}{2n}$. In fact, keeping the higher order terms only makes it more negative so that the sum in the exponent diverges to negative infinity (as the harmonic sequence does).

Thus

$$\prod_{n=1}^\infty \frac{2n-1}{2n} = \exp( -\infty ) = 0$$

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