TITLE: To show for following sequence $\lim_{n \to \infty} a_n = 0$ where $a_n$ = $1.3.5 ... (2n-1)\over 2.4.6...(2n)$
QUESTION [6 upvotes]: How can I show 
$\lim_{n \to \infty} a_n = 0$
$a_n = {1.3.5 ... (2n-1)\over 2.4.6...(2n)}$
I have shown that $a_n$ is monotonically decreasing. I thought to shown sequence is bounded from below then it automatically would converge and hence my question will be solve. But I'm unable to show its boundedness... Or there maybe another method to prove this. Thanks

REPLY [10 votes]: $\forall k\in\mathbb{N}$
$k^2+2k<k^2+2k+1\Rightarrow k(k+2)<(k+1)^2\Rightarrow\frac{k}{k+1}<\frac{k+1}{k+2}$

$\therefore a_n=\frac{1\cdot3\cdot5...(2n-1)}{2\cdot4\cdot6...2n}=\frac{1}{2}\cdot\frac{3}{4}...\frac{k}{k+1}...\frac{2n-1}{2n}<\frac{2}{3}\cdot\frac{4}{5}...\frac{k+1}{k+2}...\frac{2n}{2n+1}=\frac{2\cdot4\cdot6...2n}{1\cdot3\cdot5...(2n+1)}$where, $k$ is an odd natural number. 
$\Rightarrow {a_n}^2<\frac{1\cdot3\cdot5...(2n-1)}{2\cdot4\cdot6...2n}\cdot\frac{2\cdot4\cdot6...2n}{1\cdot3\cdot5...(2n+1)}=\frac{1}{2n+1}$

$\Rightarrow 0<a_n<\frac{1}{\sqrt{2n+1}}$

Using Squeeze theorem, we get $\underset{n\rightarrow\infty}{\lim} a_n=0$.