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 ($\frac{2\cdot4\cdot\ldots\cdot2n}{2\cdot4\cdot\ldots\cdot2n}$)
 A: Simple Answer
The product on the right of the "$\cdot$" is bounded above by $1$, so we get
$$
\lim_{n\to\infty}\frac1n\,\overbrace{\prod_{k=1}^n\frac{2k-1}{2k}}^{\le1}=0\tag1
$$

Better Bounds
The fact that the product is less than $1$ is enough to prove that the expression tends to $0$ as $n\to\infty$. However, we can get a much better bound using the inequalities
$$
\sqrt{\frac{2k-2}{2k}}\le\frac{2k-1}{2k}\le\sqrt{\frac{2k-1}{2k+1}}\tag2
$$
which can be proven by squaring and cross-multiplying. Using $(2)$, we get
$$
\frac1{2\sqrt{n}}=\frac12\prod_{k=2}^n\sqrt{\frac{2k-2}{2k}}\le\prod_{k=1}^n\frac{2k-1}{2k}\le\prod_{k=1}^n\sqrt{\frac{2k-1}{2k+1}}=\frac1{\sqrt{2n+1}}\tag3
$$
$(3)$ suggests a more interesting question: find the value of
$$
\frac12\le\lim_{n\to\infty}\sqrt{n}\,\prod_{k=1}^n\frac{2k-1}{2k}\le\frac1{\sqrt2}\tag4
$$

Series Rather than Sequence
Perhaps the question was to compute $\sum\limits_{n=1}^\infty a_n$? That sum converges and has a simple closed form.
A: I will rewrite $a_{n}$ for clarity.
$$a_{n}=\frac{1}{n}\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\cdot\dots\cdot\left(\frac{2n-1}{2n}\right).$$
Notice that the biggest term in this product is $\frac{2n-1}{2n}$ (except for when $n=1$). This means that if we multiply $\frac{2n-1}{2n}$ by itself $n$ times, then
$$0\le a_{n}=\frac{1}{n}\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\cdot\dots\cdot\left(\frac{2n-1}{2n}\right)\le b_{n}=\frac{1}{n}\left(\frac{2n-1}{2n}\right)^{n}.$$
Taking the limit of the right side, we have
$$\lim_{n\to\infty}b_{n}=\lim_{n\to\infty}\frac{1}{n}\left(\frac{2n-1}{2n}\right)^{n}=\lim_{n\to\infty}\frac{1}{n}\cdot\lim_{n\to\infty}\left(\frac{2n-1}{2n}\right)^{n}.$$
To evaluate $\lim_{n\to\infty}\left(\frac{2n-1}{2n}\right)^{n}$, notice that it is equivalent to
$$\lim_{n\to\infty}e^{\ln\left(\left(\frac{2n-1}{2n}\right)^{n}\right)}=\lim_{n\to\infty}e^{n\ln\left(1-\frac{1}{2n}\right)}=\lim_{n\to\infty}e^{\frac{\ln\left(1-\frac{1}{2n}\right)}{1/n}}=e^{\lim_{n\to\infty}\frac{\ln\left(1-\frac{1}{2n}\right)}{1/n}}$$
Using L'Hospital's rule, we get
$$e^{\lim_{n\to\infty}\frac{1/\left(2n^{2}\right)}{-1/n^{2}}}=e^{\lim_{n\to\infty}-\frac{1}{2}}=\frac{1}{\sqrt{e}}.$$
Thus,
$$\lim_{n\to\infty}\frac{1}{n}\left(\frac{2n-1}{2n}\right)^{n}=\lim_{n\to\infty}\frac{1}{n}\frac{1}{\sqrt{e}}=0.$$
But remember that this is our upper-bound for $a_{n}$, and so by the Squeeze theorem,
$$\lim_{n\to\infty}0\le\lim_{n\to\infty}a_{n}\le\lim_{n\to\infty}b_{n}\implies0\le\lim_{n\to\infty}a_{n}\le0.$$
Thus, the limit is $0$.
A: 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
$m \to \infty$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
a_{n} & \equiv{1 \over n}\,
{1\cdot3\cdot\ldots\cdot\pars{2n - 1} \over 2\cdot4\cdot\ldots\cdot\pars{2n}} =
{1 \over n}\,{\prod_{k = 1}^{n}\pars{2k - 1} \over
\prod_{q = 1}^{n}\pars{2q}}
\\[5mm] & ==
{1 \over n}\,{2^{n}\prod_{k = 1}^{n}\pars{k - 1/2} \over
2^{n}\prod_{q = 1}^{n}q} =
{1 \over n}\,{\pars{1/2}^{\overline{n}} \over n!} =
{1 \over n}\,{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!}
\\[5mm] & =
{1 \over n}\,
{\pars{n - 1/2}! \over n!}\,{1 \over \root{\pi}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{\pi^{-1/2} \over n}\,
{\root{2\pi}\pars{n - 1/2}^{n}\expo{-\pars{n - 1/2}} \over
\root{2\pi}n^{n + 1/2}\expo{-n}}
\\[5mm] & =
{\pi^{-1/2} \over n}\,
{n^{n}\bracks{1 - \pars{1/2}/n}^{n}\expo{1/2} \over
n^{n + 1/2}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{1 \over \root{\pi}}\,{1 \over n^{3/2}}
\\[5mm] & 
\stackrel{\mrm{as}\ n\ \to\ \infty}{\Large\to}\,\,\,\bbx{0}
\end{align}

Note that $\ds{\Gamma\pars{1/2} = \root{\pi}}$.

