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.

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 $\color{red}{\sqrt[3]{4}}$.
