Convergence of $a_n = \frac{1}{n} \cdot \frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)}$
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.
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 $0$.