How to compute the value of $1 \cdot \frac{2}3 \cdot \frac 45 \cdots$, as $n$ goes to infinity? It's supposed to be $\dfrac 12\sqrt{\dfrac \pi n}$.
 A: Hint.
$$
1\cdot\frac{2}{3}\cdot\frac{4}{5}\cdots\frac{2n}{2n+1}=\frac{2^{2n+1} n!(n+1)!}{(2n+2)!}
$$
Now we use Stirling's asymptotic approximation
$$
\lim_{n\to\infty}\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n}=1,
$$
and obtain
$$
\frac{2^{2n+1} n!(n+1)!}{(2n+2)!}=\frac{\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n}
\cdot\frac{(n+1)!}{\sqrt{2\pi (n+1)}\left(\frac{n+1}{\mathrm{e}}\right)^{n+1}}}{\frac{(2n+2)!}{\sqrt{2\pi (2n+2)}\left(\frac{2n+2}{\mathrm{e}}\right)^{2n+2}}}\cdot
\frac{2^{2n+1}\sqrt{2\pi (2n+2)}\left(\frac{2n+2}{\mathrm{e}}\right)^{2n+2}}{\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n\cdot\sqrt{2\pi (n+1)}\left(\frac{n+1}{\mathrm{e}}\right)^{n+1}}.
$$
The first factor of the right-hand side tends to $1$.
A: $\newcommand{\+}{^{\dagger}}%
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$$
\half\,{2 \over 3}\,{4 \over 5}\cdots{2n \over 2n + 1}
=
{\pars{1.2.4\ldots 2n}^{2}/4 \over 2.3.4.5\ldots\pars{2n}\pars{2 n + 1}}
={1 \over 4}\,{\pars{2\ n!}^{2} \over \pars{2n + 1}!}
$$

When $n \gg 1$:
  \begin{align}
&\half\,{2 \over 3}\,{4 \over 5}\cdots{2n \over 2n + 1}
\sim{\pars{\root{2\pi}n^{n + 1/2}\expo{-n}}^{2}
  \over \root{2\pi}\pars{2n + 1}^{2n + 3/2}\expo{-\pars{2n + 1}}}
={{\rm e} \over \root{2\pi}}\,{n^{2n + 1} \over \pars{2n + 1}^{2n + 3/2}}
\\[3mm]&={{\rm e} \over \root{2\pi}}\,{1 \over 2^{2n + 1}\root{2n + 1}}\,
\pars{1 + {1 \over 2n}}^{-2n - 1}
\\[3mm]&\sim{{\rm e} \over \root{2\pi}}\,{1 \over 2^{2n + 1}\root{2n + 1}}\,
\exp\pars{-\bracks{2n + 1}\,{1 \over 2n}}
\sim{1 \over \root{2\pi}}\,{1 \over 2^{2n + 1}\root{2n + 1}}
\end{align}

$$
\color{#00f}{\large%
\lim_{n \to \infty}\pars{\half\,{2 \over 3}\,{4 \over 5}\cdots{2n \over 2n + 1}}}
=\lim_{n \to \infty}
\bracks{{1 \over \root{2\pi}}\,{1 \over 2^{2n + 1}\root{2n + 1}}}
=\color{#00f}{\large 0}
$$
A: From the Wallis's product in the following form ( Advanced Calculus by Angus Taylor, formula (20.7-9)) 
\begin{equation*}
\frac{\pi }{2}=\lim_{n\rightarrow \infty }\left( \frac{2\cdot 4\cdots (2n)}{
1\cdot 3\cdots (2n-1)}\right) ^{2}\frac{1}{2n+1}\tag{1}
\end{equation*}
follows that the limit in the title is zero. 
ADDED. The formula $(1)$ is a consequence of the following double inequality
\begin{equation*}
\frac{2n}{2n+1}\frac{\pi }{2}<\left( \frac{2\cdot 4\cdots (2n)}{1\cdot
3\cdots (2n-1)}\right) ^{2}\frac{1}{2n+1}<\frac{\pi }{2},\tag{2}
\end{equation*}
which can be derived integrating over the interval $]0,\pi/2[$
$$\sin^{2n+1}\theta<\sin^{2n}\theta<\sin^{2n-1}\theta.\tag{3}$$
Inequality $(2)$ can be rewritten as
\begin{equation*}
\frac{\sqrt{n}}{2n+1}\sqrt{\pi }<\frac{2\cdot 4\cdots (2n)}{1\cdot 3\cdots
(2n-1)\left( 2n+1\right) }<\sqrt{\frac{\pi }{2\left( 2n+1\right) }}<\frac{1}{
2}\sqrt{\frac{\pi }{n}},\tag{4}
\end{equation*}
thus proving the asymptotic expression you indicate in the question body. Is this what you mean?
A: Try this transformation:
$$ R_n = \frac{1\cdot2\cdot4\cdots(2n)}{1\cdot3\cdot5\cdots(2n+1)} = \frac{(1\cdot2\cdot4\cdots(2n))^2}{1\cdot2\cdot3\cdot4\cdot5\cdots(2n)(2n+1)} = \frac{(2^n n!)^2}{(2n+1)!}$$
and then use Sterling approximation to expand the factorials.
