Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$ We can find the sum of infinite geometric series but I am stuck on this problem.
Find the sum of the following infinite series:
$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots$$
 A: Using binomial expansion, we have: 
$(1-x)^{-2/3} = 1 + \dfrac{\frac{2}{3}}{1!}x + \dfrac{\frac{2}{3} \cdot \frac{5}{3}}{2!} x^2 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3}}{3!}x^3 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3} \cdot \frac{11}{3}}{3!}x^4  + \cdots$
$(1-x)^{-2/3} = 1 + \dfrac{2}{3}x + \dfrac{2 \cdot 5}{3 \cdot 6} + \dfrac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}x^3 + \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}x^4  + \cdots$
Plug in $x = \dfrac{1}{2}$ to get: 
$(1-\frac{1}{2})^{-2/3} = 1 + \dfrac{2}{3}\cdot\dfrac{1}{2} + \dfrac{2 \cdot 5}{3 \cdot 6}\cdot\dfrac{1}{2^2} + \dfrac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\cdot\dfrac{1}{2^3} + \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\cdot\dfrac{1}{2^4}  + \cdots$
$2^{2/3} = 1 + \dfrac{2}{6} + \dfrac{2 \cdot 5}{6 \cdot 12}+ \dfrac{2 \cdot 5 \cdot 8}{6 \cdot 12 \cdot 18}+ \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24}  + \cdots$
Finally, subtract $1$ and divide both sides by $2$ to get: 
$\dfrac{2^{2/3} - 1}{2} = \dfrac{1}{6} + \dfrac{5}{6 \cdot 12}+ \dfrac{ 5 \cdot 8}{6 \cdot 12 \cdot 18}+ \dfrac{ 5 \cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24}  + \cdots$
A: $$\begin{align}
y  &= \sum_{k=1} \frac{\prod_{j=2}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{A}
\\ &= \left(-\frac 12\right) + \sum_{k=0}
   \frac{\left(-\frac 12\right)\prod_{j=0}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{B}
\\ &= \left(-\frac 12\right) + \sum_{k=0}
   \frac{\left(-\frac 13\right)^{k+1}\left(-\frac 12\right)\prod_{j=0}^k 1/3 - j} {6^k k!}\tag{C}
\\ &= \left(-\frac 12\right) + \left(-\frac 13\right)\left(-\frac 12\right)\sum_{k=0} 
   \frac{\prod_{j=0}^k 1/3 - j} {k!}\left(-\frac 1 {3\cdot 6}\right)^k\tag{D}
\\ &= \left(-\frac 12\right) + \left(\frac 16\right) \sum_{k=0} \frac{\prod_{j=0}^k 1/3 - j} {k!}\left(-\frac 1 {18}\right)^k\tag{E}
\end{align}$$
Generalized binomial theorem is:
$$(1 + x)^n = \sum_{k=0} \frac{ \prod_{j=0}^k n - j } {k!} x^k\tag{F}$$
So if I don't have a type then
$$y = \left(-\frac 12\right) + \left(\frac 16\right)\left(1 + -\frac{1}{18}\right)^{1/3} = \frac{1}{18}\sqrt[3]{\frac{17}{2}} - \frac 12\tag{G}$$
A: $$\frac16+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots=$$
$$=\frac12\cdot\bigg[\frac26+\frac{2\cdot5}{6\cdot12}+\frac{2\cdot5\cdot8}{6\cdot12\cdot18}+\frac{2\cdot5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots\bigg]=$$
$$=\frac12\cdot\bigg[\frac{(3-1)}{(6\cdot1)}+\frac{(3-1)\cdot(6-1)}{(6\cdot1)\cdot(6\cdot2)}+\frac{(3-1)\cdot(6-1)\cdot(9-1)}{(6\cdot1)\cdot(6\cdot2)\cdot(6\cdot3)}+\dots\bigg]=$$
$$=\frac12\cdot\bigg[\frac{(3\cdot1-1)}{(6\cdot1)}+\frac{(3\cdot1-1)\cdot(3\cdot2-1)}{(6\cdot1)\cdot(6\cdot2)}+\frac{(3\cdot1-1)\cdot(3\cdot2-1)\cdot(3\cdot3-1)}{(6\cdot1)\cdot(6\cdot2)\cdot(6\cdot3)}+\dots\bigg]=$$
$$=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\prod_{k=1}^n(3k-1)}{6^n\cdot n!}=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\prod_{k=1}^n\bigg(k-\frac13\bigg)}{2^n\cdot n!}=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\bigg(-\frac13\bigg){\Large!}\cdot\prod_{k=1}^n\bigg(k-\frac13\bigg)}{\bigg(-\dfrac13\bigg){\Large!}\cdot2^n\cdot n!}=$$
$$=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\bigg(n-\frac13\bigg){\Large!}}{\bigg(-\dfrac13\bigg){\Large!}\cdot2^n\cdot n!}=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\bigg(n-\frac13\bigg){\Large!}}{\bigg(-\dfrac13\bigg){\Large!}\cdot n!}\cdot\bigg(\frac12\bigg)^n=\frac12\cdot\sum_{n=1}^\infty{n-\frac13\choose n}\bigg(\frac12\bigg)^n$$
$$=\frac12\cdot\sum_{n=1}^\infty{\frac13-1\choose n}\bigg(-\frac12\bigg)^n=\frac12\cdot\bigg[-1+\sum_{n={\color{red}0}}^\infty{-\frac23\choose n}\bigg(-\frac12\bigg)^n\bigg]=$$
$$=\frac12\cdot\bigg[-1+\bigg(1-\frac12\bigg)^{^{-\tfrac23}}\bigg]=-\frac12+\frac1{\sqrt[3]2}.$$
