Alternate solution, January 18, 2007

$\frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} = \prod_{k=1}^n \frac{2k-1}{2k} = \prod_{k=1}^n \left(1 - \frac{1}{2k}\right) <$

$\prod_{k=1}^n \exp \left(-1/(2k)\right) = \exp \left(-\sum_{k=1}^n \frac{1}{2k} \right) .$

But $\lim_{n\to\infty} \exp\left(-\sum_{k=1}^n \frac{1}{2k}\right) = 0$ , since $\sum_{k=1}^\infty \frac{1}{2k} = \infty$ .

Therefore $\lim_{n\to\infty} \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} = 0 .$

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