Find sum of series $frac{1}{6} +frac{5}{6cdot12} +frac{5cdot8}{6cdot12cdot18} +frac{5cdot8cdot11}{6cdot12cdot18cdot24}+...$

How to find sum of above series

$$\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...$$
How to find sum of series I can find its convergence but not sum of series.

Can anyone explain?

Answer

We want to compute:

$$S = \sum_{n\geq 1}\frac{1}{6^n n!}\cdot\frac{1}{2}\prod_{k=1}^{n}(3k-1) = \sum_{n\geq 1}\frac{\Gamma\left(n+\frac{2}{3}\right)}{2^{n+1}\,\Gamma\left(\frac{2}{3}\right)\,\Gamma(n+1)}=\sum_{n\geq 1}\frac{B\left(n+\frac{2}{3},\frac{1}{3}\right)}{2^{n+1}\cdot\frac{2\pi}{\sqrt{3}}}$$
that by Euler's Beta function equals
$$\frac{\sqrt{3}}{4\pi}\sum_{n\geq 1}\int_{0}^{1}\frac{1}{2^n} x^{n-1/3}(1-x)^{-2/3}\,dx =\frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{x^{2/3}}{(1-x)^{2/3}(2-x)}\,dx$$
or

$$\frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{(1-x)^{2/3}}{x^{2/3}(1+x)}\,dx =\frac{3\sqrt{3}}{4\pi}\int_{0}^{1}\frac{(1-x^3)^{2/3}}{1+x^3}\,dx=\color{red}{\frac{2^{2/3}-1}{2}}.$$