Limit of a sequence of products How do you prove the following?
$$\lim_{n\,\to\,\infty}\,\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\ =\ 0$$

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.
