TITLE: Sum $\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$
QUESTION [5 upvotes]: A series is given as follows 
$$\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$$
Can you give me hints to get started finding its value? Thanks.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&{1 \over 6} + {5 \over 6 \cdot 12} + {5 \cdot 8 \over 6 \cdot 12 \cdot 18} + {5 \cdot 8 \cdot 11 \over 6\cdot 12\cdot 18\cdot 24} + \cdots =
{1 \over 6} + \sum_{n = 0}^{\infty}{\prod_{k = 0}^{n}\pars{3k + 5} \over \prod_{k = 0}^{n + 1}\pars{6k + 6}}
\\[5mm] = &\
{1 \over 6} + \sum_{n = 0}^{\infty}{3^{n + 1}\prod_{k = 0}^{n}\pars{k + 5/3} \over 6^{n + 2}\prod_{k = 0}^{n + 1}\pars{k + 1}} =
{1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}\,{\pars{5/3}^{\overline{n + 1}} \over \pars{n + 2}!}
\\[5mm] = &\
{1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}\,
{\Gamma\pars{n + 8/3}/\Gamma\pars{5/3} \over \pars{n + 2}!} =
{1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}\,
{\pars{n + 5/3}! \over \pars{2/3}!\pars{n + 2}!}
\\[5mm] = &\
{1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}
{n + 8/3 \choose n + 2}{1 \over n + 8/3} =
{1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}
{-5/3 \choose n + 2}\pars{-1}^{n + 2}\,{1 \over n + 8/3}
\\[5mm] = &\
{1 \over 6} + {1 \over 6}\sum_{n = 2}^{\infty}{1 \over 2^{n - 1}}
{-5/3 \choose n}\pars{-1}^{n}\,{1 \over n + 2/3} =
{1 \over 6} + {1 \over 6}\sum_{n = 2}^{\infty}{\pars{-1}^{n} \over 2^{n - 1}}
{-5/3 \choose n}\int_{0}^{1}x^{n - 1/3}\,\dd x
\\[5mm] = &\
{1 \over 6} + {1 \over 3}\int_{0}^{1}
\sum_{n = 2}^{\infty}{-5/3 \choose n}\pars{-\,{x \over 2}}^{n}
\,{\dd x \over x^{1/3}}
=
{1 \over 6} + {1 \over 3}\ \underbrace{\int_{0}^{1}
\bracks{\pars{1 - {x \over 2}}^{-5/3} - 1 - {5 \over 3}\,{x \over 2}}
\,{\dd x \over x^{1/3}}}_{\ds{-2 + {3 \over 2^{1/3}}}}
\\[5mm] = &\
\bbx{{1 \over 2^{1/3}} - {1 \over 2}} \approx 0.2937
\end{align}