Convergence of $\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(\frac{2}{3}+n)}{(1+\frac{3}{2})\ldots(\frac{3}{2}+n)}.$ 
Examine the convergence of the series:
$$\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}.$$

Attempt. Using ratio test for $a_n:=\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}>0$, we have:
$$\frac{a_{n+1}}{a_n}=\frac{\frac{2}{3}+n+1}{\frac{3}{2}+n+1}\to 1,$$
so we do not conclude. Also $\sqrt[n]{(1+a)\ldots(n+a)}\to +\infty$ for $a>0$, so the root test also is inconclusive.
Thanks for the help.

Let us use Gauss's test.
$$\frac{a_n}{a_{n+1}} = \frac{n+\frac52}{n+\frac53} = 1 + \frac{\frac56}{n(1+\frac{5}{3n})} = 1 + \frac{\alpha}{n} + \underline{O}(n^{-2})$$
with $\alpha = \frac{5}{6} < 1$. Hence $\sum_n a_n$ diverges.
