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:-
I can not figure out the general term of this series. It is looking like a power series as follows:-
So how to solve it and is there any easy way to find out the general term of such type of series?

Solutions Collecting From Web of "Finding the sum of the infinite series whose general term is not easy to visualize: $\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots$"

Notice $$
& \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]
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)


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? $

This is $\frac{1}{2} \left(_2F_1(2/3,1;1;\frac{1}{2}) – 1\right)$ where $_2F_1$ is the hypergeometric function. For these parameters this evaluates to $\frac{1}{2} \left(_2F_1(2/3,1;1;\frac{z}{2}) – 1\right) = \frac{1}{2} \left( (1 – \frac{z}{2})^{-2/3} – 1\right)$.

Let us consider $$\Sigma=\frac{1}{6}+\frac{5}{6\times 12}+\frac{5\times8}{6\times12\times18}+\frac{5\times8\times11}{6\times12\times18\times24}+\cdots$$ and let us rewrite it as $$\Sigma=\frac{1}{6}+\frac 16\left(\frac{5}{ 12}+\frac{5\times8}{12\times18}+\frac{5\times8\times11}{12\times18\times24}+\cdots\right)=\frac{1}{6}+\frac 16 \sum_{n=0}^\infty S_n$$ using $$S_n=\frac{\prod_{i=0}^n(5+3i)}{\prod_{i=0}^n(12+6i)}$$ Using the properties of the gamma function, we have $$\prod_{i=0}^n(5+3i)=\frac{5\ 3^n \Gamma \left(n+\frac{8}{3}\right)}{\Gamma \left(\frac{8}{3}\right)}$$ $$\prod_{i=0}^n(12+6i)=6^{n+1} \Gamma (n+3)$$ which make $$S_n=\frac{5\ 2^{-n-1} \Gamma \left(n+\frac{8}{3}\right)}{3 \Gamma
\left(\frac{8}{3}\right) \Gamma (n+3)}$$ $$\sum_{n=0}^\infty S_n=\frac{10 \left(3\ 2^{2/3}-4\right) \Gamma \left(\frac{2}{3}\right)}{9 \Gamma
\left(\frac{8}{3}\right)}=3\ 2^{2/3}-4$$ $$\Sigma=\frac{1}{\sqrt[3]{2}}-\frac{1}{2}$$

It seems like (without knowing the exact form we can not proceed) the double of $n$th term of your series is given by $$u_n=\dfrac{\cdots(3n-1)}{\cdots(6n)}.$$ 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 $\sum^{n}_{r=1} u_r$.