Find sum of 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 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?
 A: Like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $ 
as $$\dfrac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}=\dfrac32\cdot\dfrac{-2/3\cdot-5/3\cdot-8/3\cdot-11/3}{4!}\left(-\dfrac36\right)^4$$
$$\dfrac16=\dfrac32\cdot\dfrac{-2/3}{1!}\left(-\dfrac36\right)^1$$
So, the sum $$=-1+\dfrac32\cdot\left(1-\dfrac36\right)^{-2/3}$$
A: Using binomial expansion
\begin{eqnarray*}
(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots
\end{eqnarray*}
with $x=-\frac{1}{2}$ and $n=-\frac{2}{3}$.
\begin{eqnarray*}
(1-\frac{1}{2})^{-\frac{2}{3}}=1+\left(-\frac{2}{3}\right)\left(-\frac{1}{2}\right)+\frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{2}\left(-\frac{1}{2}\right)^2+\frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)\left(-\frac{8}{3}\right)}{3!}\left(-\frac{1}{2}\right)^3+\cdots
\end{eqnarray*}
Rearrange a bit & we have 
\begin{eqnarray*}
\color{red}{\frac{2^{\frac{2}{3}}-1}{2}}=\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\cdots
\end{eqnarray*}
A: 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 
$$ \frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{(1-x)^{2/3}}{x^{2/3}(1+x)}\,dx =\frac{3\sqrt{3}}{4\pi}\int_{0}^{1}\frac{(1-x^3)^{2/3}}{1+x^3}\,dx=\color{red}{\frac{2^{2/3}-1}{2}}.$$
