<s>[INST] Convergence of $a_n = \frac{1}{n} \cdot \frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)}$ I am not finding any path to solve it. Online calculators are not helping either... The answer is it converges and converges to 0.
I believe it is the same as the result of
$$\lim_{x\to\infty}\frac1x\times\frac{(2x)!}{4^x\times(x!)^2}$$
I got the idea above by multiplying $a_n$ by ($\frac{2\cdot4\cdot\ldots\cdot2n}{2\cdot4\cdot\ldots\cdot2n}$)
 [/INST] If
$a_n 
= \frac{1}{n} \cdot \frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)}
= \frac{\prod_{k=1}^n(2k-1)}{n\prod_{k=1}^n(2k)}
$
then
$a_{n+1}
= \frac{\prod_{k=1}^{n+1}(2k-1)}{(n+1)\prod_{k=1}^{n+1}(2k)}
$
so
$\begin{array}\\
\frac{a_n}{a_{n+1}}
&=\frac{(n+1)2(n+1)}{n(2n+1)}\\
&=\frac{2(n^2+2n+1)}{2n^2+n}\\
&=\frac{n^2+2n+1}{n^2+n/2}\\
&=\frac{n^2+n/2+(3/2)n+1}{n^2+n/2}\\
&=1+\frac{(3/2)n+1}{n(n+1/2)}\\
&=1+\frac{(3/2)(n+1/2)+1/4}{n(n+1/2)}\\
&=1+\frac{3}{2n}+\frac{1}{4n(n+1/2)}\\
&>1+\frac{3}{2n}\\
\end{array}
$
Since,
if $c_k > 0$,
then
$\prod_{k=1}^n (1+c_k)
\ge 1+\sum_{k=1}^n c_k
$ 
(easily proved by induction),
$\begin{array}\\
\frac{a_1}{a_{m+1}}
&=\prod_{n=1}^m\frac{a_n}{a_{n+1}}\\
&>\prod_{n=1}^m(1+\frac{3}{2n})\\
&\gt 1+\sum_{n=1}^m\frac{3}{2n}\\
&\gt \frac32\sum_{n=1}^m\frac1{n}\\
&\gt \frac32\ln(m)
\qquad\text{(well-known harmonic sum)}\\
\end{array}
$
so
$a_{m+1}
\lt \frac{2a_1}{3\ln(m)}
= \frac{1}{3\ln(m)}
\to 0$
as
$m \to \infty$.
</s>