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?

It seems like (without knowing the exact form we can not proceed) the double of $n$th term of your series is given by $$u_n=\dfrac{2.5.8.11\cdots(3n-1)}{6.12.18.24\cdots(6n)}.$$ By ratio test this series is convergent and we can find the exact formula for $n$th partial summation $\sum^{n}_{r=1} u_r$ by observing $$(6n+6)u_{n+1}=u_n(3n+2)\forall n\in\mathbb{N}.$$ Find a function $f$ with the property $$f(r+1)=f(r)\left(\dfrac{3r+2}{6r+6}\right)$$ (in terms of $u_r$) and $$u_r=f(r)-f(r+1).$$ Then Telescope to find $\sum^{n}_{r=1} u_r$.
