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.

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:
$$
\sum^\infty_{n=0}\left(\!
    \begin{array}{c}
      n-\frac{1}{3} \\
      -\frac{1}{3}
    \end{array}
  \!\right)2^{-n}=\left(1-\frac{1}{2}\right)^{-\frac{2}{3}}=2^{\frac{2}{3}}
$$
