$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$ How to find the sum 

$$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24}  + \cdots.$$

I have tried to make this series in  the form $1 + n x  + \frac {n (n + 1) x^2}{2!} + \frac {n (n+1) (n+2)x^3}{3!} +  \cdots$. But I failed to do so. Can anyone please help me
 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? $,
If $\displaystyle S=\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24}  + \cdots.$
$$2S=\frac {2}{6} + \frac {2\cdot5}{6.12} + \frac {2\cdot5. 8}{6.12.18} + \frac {2\cdot5.8.11}{6.12.18.24}  + \cdots$$
$$2S+1+\dfrac13$$
$$=1+\dfrac{-\dfrac23}{1!}\left(-\dfrac36\right)+\dfrac{-\dfrac23\left(-\dfrac23-1\right)}{2!}\left(-\dfrac36\right)^2+\dfrac{-\dfrac23\left(-\dfrac23-1\right)\left(-\dfrac23-2\right)}{3!}\left(-\dfrac36\right)^3+ \cdots$$
$$=\left(1-\dfrac12\right)^{-2/3}$$
A: Let $a_n=\frac{1}{2}\prod_{k=1}^{n}(3k-1)$ and $b_n=\prod_{k=1}^{n}(6k)$. Then
$$ \frac{a_n}{b_n} = \frac{\Gamma\left(n+\frac{2}{3}\right)}{2^{n+1}\Gamma(n+1)\Gamma\left(\frac{2}{3}\right)}=\frac{\pi}{2^n \sqrt{3}} B\left(n+\frac{2}{3},\frac{1}{3}\right)=\frac{\pi}{\sqrt{3}}\int_{0}^{1}\frac{x^{n-1/3}(1-x)^{-2/3}}{2^n}\,dx $$
hence
$$ \sum_{n\geq 1}\frac{a_n}{b_n} = \frac{\pi}{\sqrt{3}}\int_{0}^{1}\frac{x^{2/3}}{(2-x)(1-x)^{2/3}}\,dx\stackrel{\frac{x}{1-x}\mapsto z}{=}\frac{\pi}{\sqrt{3}}\int_{0}^{+\infty}\frac{z^{2/3}}{(z+1)(z+2)}\,dz. $$
By setting $z=v^3$ and applying partial fraction decomposition we get
$$ \sum_{n\geq 1}\frac{a_n}{b_n}=\color{red}{\frac{2^{2/3}-1}{2}}. $$
The same can be proved by considering the Maclaurin series of $(1-x)^{-2/3}$ evaluated at $x=\frac{1}{2}$.
