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....
 A: List out your $a_n$ clearly.$$a_n =\prod_{i=1}^n \left(\frac{2i-1}{3i-2}\right)$$
$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\lim_{n \to \infty}\frac{2(n+1)-1}{3(n+1)-2}=\frac{2}{3}$$
A: 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.
A: Hint:
$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{1\cdot3\cdot5\cdots\cdot(2n-1)\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=2n+1$$
and not
$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{2n+1}{2n-1}.$$
A: $$a_k=\prod_{k=1}^{n}\frac{2k-1}{3k-2}=\left(\frac{2}{3}\right)^n\cdot\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\cdot B\left(n+\tfrac{1}{2},\tfrac{1}{6}\right)\tag{1}$$
implies
$$\begin{eqnarray*} \sum_{k\geq 1}a_k &=& \frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{1}\sum_{n\geq 1}\left(\frac{2}{3}\right)^n (1-x)^{-5/6} x^{n-1/2}\,dx \\&=&\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{\pi/2}\frac{4(1-\cos^2\theta)}{\cos^{2/3}\theta(1+2\cos^2\theta)}\,d\theta\\&=&\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{1}12\sqrt{1-z^6}\,\frac{dz}{1+2z^6}\tag{2}\end{eqnarray*}$$
hence trivially
$$ \sum_{k\geq 1}a_k = \tfrac{1}{2}\cdot\phantom{}_2 F_1\left(1,\tfrac{3}{2};\tfrac{5}{3};\tfrac{2}{3}\right) \leq \frac{3}{\pi}2^{4/3}\,\Gamma\left(\tfrac{2}{3}\right)^2.\tag{3}$$
In general, $\phantom{}_{p+1}F_p(\ldots; z)$ is convergent for any $|z|<1$.
