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:
$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots$$

$$\begin{align}
y  &= \sum_{k=1} \frac{\prod_{j=2}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{A}
\\ &= \left(-\frac 12\right) + \sum_{k=0}
   \frac{\left(-\frac 12\right)\prod_{j=0}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{B}
\\ &= \left(-\frac 12\right) + \sum_{k=0}
   \frac{\left(-\frac 13\right)^{k+1}\left(-\frac 12\right)\prod_{j=0}^k 1/3 - j} {6^k k!}\tag{C}
\\ &= \left(-\frac 12\right) + \left(-\frac 13\right)\left(-\frac 12\right)\sum_{k=0} 
   \frac{\prod_{j=0}^k 1/3 - j} {k!}\left(-\frac 1 {3\cdot 6}\right)^k\tag{D}
\\ &= \left(-\frac 12\right) + \left(\frac 16\right) \sum_{k=0} \frac{\prod_{j=0}^k 1/3 - j} {k!}\left(-\frac 1 {18}\right)^k\tag{E}
\end{align}$$
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
$$y = \left(-\frac 12\right) + \left(\frac 16\right)\left(1 + -\frac{1}{18}\right)^{1/3} = \frac{1}{18}\sqrt[3]{\frac{17}{2}} - \frac 12\tag{G}$$
