Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$ How do I begin to evaluate this limit: $$\lim_{n\to \infty}\ \frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}\;?$$
Thanks a lot.

A simple, but famous trick works here: Observe that $(n-1)(n+1) = n^2 - 1 \leq n^2$. Thus we have 
$$ \begin{align*}
& \left[ \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right]^2 \\
&= \frac{1 \cdot 1}{2 \cdot 2}\cdot\frac{3 \cdot 3}{4 \cdot 4}\cdot\frac{5 \cdot 5}{6 \cdot 6}\cdots\frac{(2n-3) \cdot (2n-3)}{(2n-2) \cdot (2n-2)}\cdot\frac{(2n-1) \cdot (2n-1)}{(2n) \cdot (2n)} \\
&= \frac{1 \cdot 3}{2 \cdot 2}\cdot\frac{3 \cdot 5}{4 \cdot 4}\cdot\frac{5 \cdot 7}{6 \cdot 6}\cdots\frac{(2n-3) \cdot (2n-1)}{(2n-2) \cdot (2n-2)}\cdot\left(\frac{2n-1}{(2n)^2}\right) \\
& \leq \frac{2n-1}{(2n)^2} \\
& \leq \frac{1}{2n}.
\end{align*} $$
Thus we have
$$\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \leq \frac{1}{\sqrt{2n}} $$
and the limit is zero. In fact, the sharp estimate
$$ \frac{1}{\sqrt{(\pi + o(1)) n}} \leq \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \leq \frac{1}{\sqrt{\pi n}} $$
holds, so the estimation above is not so far from the truth.
