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.

Solutions Collecting From Web of "Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$"

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.

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,

$${\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n – 1} \right)!!}}} \right]^2}\left( {1 – \frac{1}{{2n + 1}}} \right)\frac{1}{n} < \pi < {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n – 1} \right)!!}}} \right]^2}\frac{1}{n}$$

$$\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n – 1} \right)!!}}} \right]\sqrt {1 – \frac{1}{{2n + 1}}} < \sqrt {n\pi } < \left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n – 1} \right)!!}}} \right]$$