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....

List out your $a_n$ clearly.$$a_n =\prod_{i=1}^n \left(\frac{2i-1}{3i-2}\right)$$
$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\lim_{n \to \infty}\frac{2(n+1)-1}{3(n+1)-2}=\frac{2}{3}$$
