<s>[INST] 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....
 [/INST] $$a_k=\prod_{k=1}^{n}\frac{2k-1}{3k-2}=\left(\frac{2}{3}\right)^n\cdot\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\cdot B\left(n+\tfrac{1}{2},\tfrac{1}{6}\right)\tag{1}$$
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 $|z|<1$.
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