# 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.
• What does $\underline{O}$ mean? Dec 23, 2020 at 9:29
• @VIVID, I meant standard "Big O notation": $f(x) = O(g(x))$ iff $|f(x)| \le M g(x)$ for $x \ge x_0$. Dec 23, 2020 at 9:40
In terms of the Gamma and Beta functions we are discussing $$\sum_{n\geq 1}\frac{\Gamma\left(n+\frac{5}{3}\right)/\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(n+\frac{5}{2}\right)/\Gamma\left(\frac{5}{2}\right)}=\underbrace{\frac{\Gamma(5/2)}{\Gamma(2/3)\Gamma(5/6)}}_{K}\sum_{n\geq 1}B(n+5/3,5/6)$$ or $$K \sum_{n\geq 1}\int_{0}^{1} x^{n+2/3}(1-x)^{-1/6}\,dx=K\int_{0}^{1}x^{5/3}(1-x)^{-7/6}\,dx$$ which is clearly divergent.