Limit of a fraction of double factorials How can we show that 
$\begin{align*}
\lim_{n\rightarrow\infty} U_n = 0
\end{align*}$
where
$\begin{align*}
U_n = \frac{(n-1)!!}{n!!}=\frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}\cdots
\end{align*}$
terminates at $\displaystyle\frac{2}{3}$ (odd) or $\displaystyle\frac{1}{2}$ (even).
At first glance I thought it was 1 because each individual multiplied fraction goes to 1 as $n\rightarrow\infty$. Mahthematica says this is not case, however.
Does it help if I write it in recurrence relation?
$\begin{align*}
U_{n+2} = \frac{n+1}{n+2}U_n\\
U_{n+1} = \frac{1}{n+1}\frac{1}{U_n}
\end{align*}$

First observe the following elementary inequality
\begin{gather*}
\frac{m}{m+1}<\sqrt{\frac{m}{m+2}},\qquad \forall m\in\mathbb{N}
\end{gather*}
where $\mathbb{N}$ denotes the set of all the positive integers.
In the case of $n$ is even, we have, by using the above inequality,
\begin{align*}
0<U_n&=\frac{1}{2}\cdot \frac{3}{4}\cdot\frac{5}{6}\cdot\cdots\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-1}{2n}\\
&<\sqrt{\frac{1}{3}}\cdot \sqrt{\frac{3}{5}}\cdot \sqrt{\frac{5}{7}}\cdot\cdots\cdot \sqrt{\frac{2n-3}{2n-1}}\cdot \sqrt{\frac{2n-1}{2n+1}}\\
&=\frac{1}{\sqrt{2n+1}}<\frac{1}{\sqrt{n}},
\end{align*}
and in the case of $n$ is odd, we have similarly
\begin{align*}
0<U_n&=\frac{2}{3}\cdot \frac{4}{5}\cdot\frac{6}{7}\cdot\cdots\cdot\frac{2n-2}{2n-1}\cdot\frac{2n}{2n+1}\\
&<\sqrt{\frac{2}{4}}\cdot \sqrt{\frac{4}{6}}\cdot \sqrt{\frac{6}{8}}\cdot\cdots\cdot \sqrt{\frac{2n-2}{2n }}\cdot \sqrt{\frac{2n }{2n+2}}\\
&=\frac{\sqrt{2}}{\sqrt{2n+2}}<\frac{1}{\sqrt{n}}.
\end{align*}
In summary, we have verified that 
\begin{gather*}
0<U_n<\frac{1}{\sqrt{n}},\qquad \forall n\in\mathbb{N}.
\end{gather*}
Then by utilizing the squeezing test and the fact that 
\begin{gather*}
\lim_{n\to\infty}\frac{1}{\sqrt{n}}=0,
\end{gather*}
we can get that 
\begin{gather*}
\lim_{n\to\infty}U_n=0.
\end{gather*}
