A question related to compute infinite summation How to solve this: $$ 1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot5\cdot8}{6\cdot12\cdot18}+\cdots $$ So, the $n^{th}$ term of the sum can be written as $$a_n=\frac{2\cdot5\cdot8\cdots (2+3(n-1))}{6\cdot12\cdot18\cdots (6n)} = \frac{2\cdot5\cdot8 \cdots (2+3(n-1))}{6^n (n!)}$$ So the above sum can be written as $$ 1+\sum_{n=1}^\infty a_n .$$  Now how should I proceed to solve this. Please help me.
 A: more than a hint...
Consider the binomial expansion of $$(1-\frac 12)^{-\frac 23}$$
A: Consider 
$$
1+\sum^\infty_{n=1}\prod^n_{i=1}\frac{2+3(i-1)}{6i}=1+\sum^\infty_{n=1}\prod^n_{i=1}\left(\frac{1}{2}-\frac{1}{6i}\right)=1+\sum^\infty_{n=1}2^{-n}\prod^n_{i=1}\left(1-\frac{1}{3i}\right)
$$
Product above can be expressed in terms of Gamma-functions:
$$
\prod^n_{i=1}\left(1-\frac{1}{3i}\right)=\frac{\Gamma(n+\frac{2}{3})
}{\Gamma(\frac{2}{3})\Gamma(n+1)}
$$
Thus, we have
$$
1+\sum^\infty_{n=1}2^{-n}\frac{\Gamma(n+\frac{2}{3})
}{\Gamma(\frac{2}{3})\Gamma(n+1)}=\sum^\infty_{n=0}2^{-n}\frac{\Gamma(n+\frac{2}{3})
}{\Gamma(\frac{2}{3})\Gamma(n+1)}
$$
By definition 
$$
\left(\!
    \begin{array}{c}
      x \\
      y
    \end{array}
  \!\right) = \frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}
$$
In our case $x=n-\frac{1}{3},\quad y=-\frac{1}{3}$. Which leads us to:
$$
\sum^\infty_{n=0}\left(\!
    \begin{array}{c}
      n-\frac{1}{3} \\
      -\frac{1}{3}
    \end{array}
  \!\right)2^{-n}
$$
Recall generating function:
$$
\frac{1}{(1-z)^{m+1}}=\sum_{n\geq 0}\left(\!
    \begin{array}{c}
      m+n \\
      m
    \end{array}
  \!\right)z^n
$$
Setting $z=\frac{1}{2}$ and $m=-\frac{1}{3}$ gives us:
$$
\sum^\infty_{n=0}\left(\!
    \begin{array}{c}
      n-\frac{1}{3} \\
      -\frac{1}{3}
    \end{array}
  \!\right)2^{-n}=\left(1-\frac{1}{2}\right)^{-\frac{2}{3}}=2^{\frac{2}{3}}
$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{\pars{a}_{m} = {\Gamma\pars{a + m} \over \Gamma\pars{a}}}$ is a
  Pochhammer Symbol. 

\begin{align}
&\color{#f00}{1 + \sum_{n = 1}^{\infty}
{\prod_{k = 1}^{n}\pars{3k - 1} \over 6^{n}\,\, n!}} =
1 + \sum_{n = 1}^{\infty}
{3^{n}\prod_{k = 1}^{n}\pars{k - 1/3} \over 6^{n}\,\, n!} =
1 + \sum_{n = 1}^{\infty}{\pars{2/3}_{n} \over 2^{n}\,\, n!}
\\[5mm] = &\
1 + \sum_{n = 1}^{\infty}
{\Gamma\pars{2/3 + n}/\Gamma\pars{2/3} \over 2^{n}\,\, \Gamma\pars{n + 1}} =
1 + {1 \over \Gamma\pars{1/3}\Gamma\pars{2/3}}\sum_{n = 1}^{\infty}
{\Gamma\pars{2/3 + n}\Gamma\pars{1/3} \over 2^{n}\,\, \Gamma\pars{n + 1}}
\\[5mm] = &\
1 + {\sin\pars{\pi/3} \over \pi}\sum_{n = 1}^{\infty}
{1 \over 2^{n}}\int_{0}^{1}x^{-1/3 +n}\,\,\,\pars{1 - x}^{-2/3}\,\,\dd x
\\[5mm] = &\
1 + {\root{3} \over 2\pi}\int_{0}^{1}x^{-1/3}\,\,\pars{1 - x}^{-2/3}\,\,
\sum_{n = 1}^{\infty}\pars{x \over 2}^{n}\,\dd x
\\[5mm] = &\
1 + {\root{3} \over 2\pi}\int_{0}^{1}x^{-1/3}\,\,\pars{1 - x}^{-2/3}\,\,
{x/2 \over 1 - x/2}\,\dd x =
1 + {\root{3} \over 2\pi}\int_{0}^{1}\pars{x \over 1 - x}^{2/3}\,\,
\,{\dd x \over 2 - x}
\end{align}

With the substitution $\ds{x = {t \over 1 + t}}$:
\begin{align}
&\color{#f00}{1 + \sum_{n = 1}^{\infty}
{\prod_{k = 1}^{n}\pars{3k - 1} \over 6^{n}\,\, n!}} =
1 + {\root{3} \over 2\pi}\ \underbrace{\int_{0}^{\infty}
{t^{2/3} \over \pars{t + 1}\pars{t + 2}}\,\dd t}
_{\ds{{2\pi \over \root{3}}\,\pars{2^{2/3} - 1}}}\ =\
\color{#f00}{2^{2/3}} \approx 1.5874\label{1}\tag{1}
\end{align}


The integral, in \eqref{1}, is trivially evaluated by an integration over a suitable contour in the complex plane.

