Convergence for series failed using "Ratio Test" $$\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n-1)}{1\cdot 4\cdot 7\cdot ...(3n-2)}$$
Using Ratio test: $$\lim_{n\rightarrow \infty}\frac{\frac{2(n+1)-1}{3(n+1)-2}}{\frac{2n-1}{3n-2}}$$
which equals to : $$\lim_{n \to \infty}\frac{6n^{2}-n-2}{6n^{2}-n-1}$$
the result for latter is q=1. seemingly this is unclear if divergent or convergent.
The answer for this is convergent....

The ratio test will work if done correctly
$$
\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
=
\lim_{n\to\infty}
\frac
{
\left(\!
{\large{\frac
{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)\,\cdot\,(2n+1)\!}
{1\,\cdot\,4\,\cdot\,7\,\cdots\,(3n-2)\,\cdot\,(3n+1)\!}
}}
\!
\right)
}
{\left(\!\large{{\frac
{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}
{1\,\cdot\,4\,\cdot\,7\,\cdots\,(3n-2)}
}}\!\!\right)}
=
\lim_{n\to\infty} \frac{2n+1}{3n+1}
=
\frac{2}{3}
$$
But here's another way . . .

Show that the sequence of fractions
$$\frac{1}{1},\frac{3}{4},\frac{5}{7},...,\frac{2n-1}{3n-2}$$
is strictly decreasing.

From that, it follows that the $n$-th term is bounded above by $\bigl({\large{\frac{3}{4}}}\bigr)^{n-1}$. 

Finish via a comparison test.
