# 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?

#### 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 \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 $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots?$

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 $$u_n=\dfrac{2.5.8.11\cdots(3n-1)}{6.12.18.24\cdots(6n)}.$$ 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 $\sum^{n}_{r=1} u_r$.