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: 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.
A: This is almost a quite famous limit.
Wallis showed that
$$\lim_{n \to \infty} \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 \frac 1 n = \pi$$
or that
$$ \frac{(2n)!!}{(2n-1)!!} \sim \sqrt{\pi n}$$
Your limit is
$$\lim_{n \to \infty} \frac{(2n-1)!!}{(2n)!!}$$
which by the above asymtotical behaviour is
$$\lim_{n \to \infty} \frac{(2n-1)!!}{(2n)!!} = \lim_{n \to \infty} \frac{1}{\sqrt{\pi n}}=0$$ 
Note you can write you expression as
$$\frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}} = \frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}}\frac{{\left( {2n} \right)!!}}{{\left( {2n} \right)!!}} = \frac{{\left( {2n} \right)!}}{{{4^n}n{!^2}}}$$
so using Stirling's approximation, one has
$$\eqalign{
  & \left( {2n} \right)! \sim {\left( {\frac{{2n}}{e}} \right)^{2n}}2\sqrt {n\pi }   \cr 
  & n{!^2} \sim {\left( {\frac{n}{e}} \right)^{2n}}2n\pi  \cr} $$
from where
$$\frac{{\left( {2n} \right)!}}{{{4^n}n{!^2}}} \sim \frac{1}{{{4^n}}}\frac{{{{\left( {\frac{{2n}}{e}} \right)}^{2n}}2\sqrt {n\pi } }}{{{{\left( {\frac{n}{e}} \right)}^{2n}}2n\pi }} = \frac{1}{{\sqrt {n\pi } }}$$
as it was previously stated.
In general, 
$${\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\left( {1 - \frac{1}{{2n + 1}}} \right)\frac{1}{n} < \pi  < {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{n}$$
$$\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]\sqrt {1 - \frac{1}{{2n + 1}}}  < \sqrt {n\pi }  < \left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]$$
A: Well known equality :
$$\frac{1}{2\sqrt{n}} \leqslant \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\leqslant \frac{1}{\sqrt{2n}}$$
Proof of right side. Having
$$\frac{1}{2}<\frac{2}{3},\frac{3}{4}<\frac{4}{5},\frac{5}{6}<\frac{6}{7},\cdots,\frac{2n-3}{2n-2}<\frac{2n-2}{2n-1},\frac{2n-1}{2n}<1$$
and multiplying we obtain
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}<\frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7} \cdots \frac{2n-2}{2n-1}$$
Now multiply both sides on $ \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}$ gives
$$ \left(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\right)^2<\frac{1}{2n}$$
and we obtain
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} <\frac{1}{\sqrt{2n}}$$
Analogical way works for left side.
