Evaluate : $1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot 5\cdot 8}{6\cdot12\cdot 18}+\ldots$ Evaluate :
$$1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot 5\cdot 8}{6\cdot12\cdot 18}+\ldots$$
thought: I have tried to manipulate the series so that the sum upto n terms , can be simplified. But I can't figure out any profitable result.
 A: It is easier to recognize a Taylor series for some function of the form $(1\pm z)^{\alpha}$, but there also is an interesting approach through Euler's Beta function.

We have
$$ \frac{1}{N!6^N}\prod_{n=1}^{N}(3n-1) = \frac{\Gamma\left(N+\frac{2}{3}\right)}{2^N \Gamma(N+1)\,\Gamma\left(\frac{2}{3}\right)}=\frac{\sqrt{3}}{\pi 2^{N+1}}B\left(N+\frac{2}{3},\frac{1}{3}\right) \tag{1}$$
hence
$$\begin{eqnarray*} \sum_{N\geq 1}\frac{1}{N!6^N}\prod_{n=1}^{N}(3n-1)&=&\frac{\sqrt{3}}{2\pi}\int_{0}^{1}\sum_{N\geq 1}\frac{(1-x)^{-2/3}x^{N-1/3}}{2^N}\,dx\\&=&\frac{\sqrt{3}}{2\pi}\int_{0}^{1}\frac{x^{2/3}}{(1-x)^{2/3}(2-x)}\,dx\\\left(\frac{x}{1-x}\mapsto w\right)\quad&=&\frac{\sqrt{3}}{2\pi}\int_{0}^{+\infty}\frac{w^{2/3}}{(1+w)(2+w)}\,dw\tag{2} \end{eqnarray*}$$
where the last integral can be easily computed by setting $w=t^3$ and applying partial fraction decomposition. The final outcome is
$$\sum_{N\geq 1}\frac{1}{N!6^N}\prod_{n=1}^{N}(3n-1) = \color{red}{2^{2/3}-1} \tag{3}$$
so the given series converges to $\color{red}{\sqrt[3]{4}}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&1 + {2 \over 6} + {2 \cdot 5 \over 6 \cdot 12} +
{2 \cdot 5 \cdot 8 \over 6 \cdot 12 \cdot 18} +\cdots =
1 + \sum_{n = 0}^{\infty}{\prod_{k = 0}^{n}\pars{3k + 2} \over
\prod_{k = 0}^{n}\pars{6k + 6}}
\\[5mm] & =
1 + \sum_{n = 0}^{\infty}{1 \over 2^{n + 1}\pars{n + 1}!}\
\overbrace{\prod_{k = 0}^{n}\pars{k + {2 \over 3}}}
^{\substack{\ds{\pars{2/3}^{\overline{n + 1}} =}\\[1.5mm]
\ds{\Gamma\pars{5/3 + n}/\Gamma\pars{2/3}}}} =
1 +
\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}
{\pars{n + 2/3}! \over \pars{n + 1}!\pars{-1/3}!}
\\[5mm] & =
1 + \sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}{n + 2/3 \choose n + 1} =
1 +
\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}{-2/3 \choose n + 1}\pars{-1}^{n + 1} =
1 +
\sum_{n = 1}^{\infty}{-2/3 \choose n}\pars{-\,{1 \over 2}}^{n}
\\[5mm] & =
1 + \braces{\bracks{1 + \pars{-\,{1 \over 2}}}^{-2/3} - {-2/3 \choose 0}} =
\bbx{2^{2/3}} \approx 1.5874
\end{align}
