Limit of a fraction of double factorials

Question

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?

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

Answer 1

Hint:

1. $\{U_n\}_{n \in \text{even}}$ and $\{U_n\}_{n\in \text{odd}}$ are decreasing, limits exist.
2. $U_n U_{n+1}=\frac{1}{n+1}$, let $n$ go to infinity.