<s>[INST] Sum of infinite series How to find sum of the following series
$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\cdots={1\over 6} + \sum_{n=1}^\infty{\Pi_{i=1}^n{3i+2}\over (n+1)!6^{n+1}}$$
Please give me some hints. Thanks in advance.
 [/INST] The general term of the series reads:
$$
    \frac{1}{2}\frac{\prod_{k=1}^n (3k-1)}{\prod_{k=1}^n 6 k} = \frac{1}{2^{n+1}} \prod_{k=1}^n \left(1-\frac{1}{3k}\right) = \frac{1}{2^{n+1}} \frac{\left(2/3\right)_n}{n!} = \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n! \cdot \Gamma\left(2/3\right)}
$$
The series thus reads:
$$
  \mathcal{S}= \sum_{n=1}^\infty \frac{1}{2^{n+1}} \frac{\left(2/3\right)_n}{n!}  = \frac{1}{\Gamma(2/3)} \sum_{n=1}^\infty \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n!}
$$
Using Euler's integral:
$$
   \Gamma\left(n+2/3\right) = \int_0^\infty t^{n-1/3} \mathrm{e}^{-t} \mathrm{d}t
$$
and interchanging the summation and the integration warranted by Tonelli's theorem:
$$\begin{eqnarray}
  \mathcal{S} &=& \frac{1}{\Gamma(2/3)} \sum_{n=1}^\infty \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n!} = \frac{1}{\Gamma(2/3)} \int_0^\infty \left(\sum_{n=1}^\infty \frac{t^{n-1/3}}{2^{n+1} n!} \right) \mathrm{e}^{-t} \mathrm{d}t \\ &=& \frac{1}{\Gamma(2/3)} \int_0^\infty \left(\frac{\exp\left(t/2\right)-1}{2 t^{1/3}} \right) \mathrm{e}^{-t} \mathrm{d}t \\ &=& \frac{1}{2\Gamma(2/3)} \left( \int_0^\infty t^{-1/3} \mathrm{e}^{-t/2} \mathrm{d}t - \int_0^\infty t^{-1/3} \mathrm{e}^{-t} \mathrm{d}t \right) \\
 &=&  \frac{1}{2\Gamma(2/3)} \left( 2^{2/3} \Gamma\left(2/3\right) - \Gamma\left(2/3\right) \int_0^\infty t^{-1/3} \mathrm{e}^{-t} \mathrm{d}t \right)  = \frac{2^{2/3}-1}{2}
\end{eqnarray}
$$
Alternatively, you might note that the series term can be written in terms of a binomial:
$$
   \frac{\left(2/3\right)_n}{n!} = \binom{2/3}{n}
$$
and hence:
$$
   \mathcal{S} = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \binom{2/3}{n} = \frac{1}{2} \left( \sum_{n=0}^\infty \frac{1}{2^n} \binom{2/3}{n}  -1 \right) 
$$
this can now be finished with the Newton's generalized binomial theorem.
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