Finding the sum of the infinite series whose general term is not easy to visualize: $\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots$ I am to find out the sum of infinite series:-
$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...............$$
I can not figure out the general term of this series. It is looking like a power series as follows:-
$$\frac{1}{6}+\frac{5}{6^2\cdot2!}+\frac{5\cdot8}{6^3\cdot3!}+\frac{5\cdot8\cdot11}{6^4\cdot4!}+.....$$
So how to solve it and is there any easy way to find out the general term of such type of series?

Notice $$
\begin{align}
  & \frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \cdots\\
= & \frac12 \left[ \frac{3-1}{6} + \frac{(3-1)(6-1)}{6^2 \cdot 2!} + \frac{(3-1)(6-1)(9-1)}{6^33!} + \cdots\right]\\
= & \frac12 \left[ \frac{\frac23}{2} + \frac{\frac23(\frac23+1)}{2^2\cdot 2!}
+ \frac{\frac23(\frac23+1)(\frac23+2)}{2^3\cdot 3!} + \cdots \right]
\end{align}
$$
The sum we want has the form
$\displaystyle\;\frac12 \sum_{k=1}^\infty \frac{\prod_{\ell=0}^{k-1}(\frac23 + \ell)}{2^k k!}$.
Compare this with the expansion
$$\frac{1}{(1-z)^\alpha} = \sum_{k=0}^\infty \frac{(\alpha)_k}{k!} z^k
\quad\text{ where }\quad
(\alpha)_k = \prod\limits_{\ell=0}^{k-1}( \alpha + \ell )$$
We find the desired sum equals to
$$\frac12 \left[\frac{1}{\left(1-\frac12\right)^{2/3}} - 1\right] = \frac12\left( 2^{2/3} - 1 \right)
\approx 
0.293700525984...
$$
Notes
After I finish this answer, I have a déjà vu feeling that I have seen this before
(more than once).
Following are some similar questions I can locate, I'm sure there are more...


*

*Sum of a Hyper-geometric series. (NBHM 2011)

*Find a closed form for this infinite sum: $ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+ \dots$

*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? $
