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?
$\begin{align*}
U_{n+2} = \frac{n+1}{n+2}U_n\\
U_{n+1} = \frac{1}{n+1}\frac{1}{U_n}
\end{align*}$
 A: 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*}
A: Hint:


*

*$\{U_n\}_{n \in \text{even}}$ and $\{U_n\}_{n\in \text{odd}}$ are decreasing, limits exist.

*$U_n U_{n+1}=\frac{1}{n+1}$, let $n$ go to infinity.

A: If $n$ is even, use the inequality $1-x\le e^{-x}$ to see
$$
0\le U_n = \frac12\frac34\frac56\cdots\frac{n-1}n\le {\textstyle\exp(-\frac12)\exp(-\frac14)\exp(-\frac16)\cdots\exp(-\frac1n)=\exp -\left(\frac12+\frac14+\cdots+\frac1n\right)}.
$$
The series $\sum_{k=\rm even}^n\frac1k$ diverges to $\infty$, so the subsequence $(U_n)_{n\in \rm even}$ tends to zero. Similarly argue that the odd-indexed terms of $(U_n)$ tend to zero. These two facts imply that $U_n\to0$.
As an aside, it is not enough to observe that the odd-indexed terms of $(U_n)$ are decreasing, the even-indexed terms are decreasing, and that $U_nU_{n+1}\to 0$. A counterexample is the sequence
$$
U_n:=\begin{cases} \frac1n&\text{$n$ even}\\
\frac12+\frac1n&\text{$n$ odd}\\
\end{cases}
$$
