TITLE: How to compute the value of $1 \cdot \frac{2}3 \cdot \frac 45 \cdots$, as $n$ goes to infinity?
QUESTION [1 upvotes]: It's supposed to be $\dfrac 12\sqrt{\dfrac \pi n}$.

REPLY [1 votes]: 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?
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