Sum $\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$ A series is given as follows 
$$\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$$
Can you give me hints to get started finding its value? Thanks.
 A: Consider the binomial expansion of $(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...$
Multiplying the series term by term by $2$
On comparing, $nx=\frac{1}{3}$ and $\frac{n(n-1)}{2}x^2=\frac{5}{3.12}$
Solving for $n,x$ we get $n=\frac{-2}{3},x=\frac{-1}{2}$
So the sum becomes $\{(\frac{1}{2})^{\frac{-2}{3}}-1\}=2^\frac{2}{3}-1=4^\frac{1}{3}-1$
since $2$ has been multiplied before divide the result by $2$
A: $$\begin{align}
\sum_{n=1}^{\infty}\frac12\prod_{i=1}^n\frac{3i-1}{6i}
&=\frac12\sum_{n=1}^{\infty}\frac1{2^nn!}\prod_{i=1}^n\frac{3i-1}{3}\\
&=\frac12\sum_{n=1}^{\infty}\frac1{(-2)^nn!}\prod_{i=1}^n\frac{1-3i}{3}\\
&=\frac12\sum_{n=1}^{\infty}\frac1{(-2)^nn!}\prod_{i=1}^n\left(\frac13-i\right)\\
&=\left[\frac12\sum_{n=1}^{\infty}\frac{x^n}{n!}\prod_{i=1}^n\left(\frac13-i\right)\right]_{x=-1/2}\\
\end{align}$$
Now note that for $n\geq1$, $\prod_{i=1}^n\left(\frac13-i\right)=f^{(n)}(0)$, where $f(x)=(x+1)^{-2/3}$. We can manipulate the expression a little further to see it as a Taylor series for $f$.
$$\begin{align}
\sum_{n=1}^{\infty}\frac12\prod_{i=1}^n\frac{3i-1}{6i}
&=\left[\frac12\sum_{n=0}^{\infty}\frac{x^n}{n!}\prod_{i=1}^n\left(\frac13-i\right)-\frac12\right]_{x=-1/2}\\
&=\left[\frac12\sum_{n=0}^{\infty}\frac{x^n}{n!}f^{(n)}(0)\right]_{x=-1/2}-\frac12\\
&=\frac12f(-1/2)-\frac12\\
&=\frac12(1/2)^{-2/3}-\frac12\\
&=\sqrt[3]{1/2}-\frac12\\
\end{align}$$
A: For any $n\in\mathbb{N}^+$ we have
$$ \prod_{k=1}^{n}(3k+2) = \frac{3^n\,\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)}\tag{1}$$
and the value of the RHS at $n=0$ equals $1$. It follows that the whole series can be represented as
$$ S=\sum_{n\geq 0}\frac{1}{6^{n+1}(n+1)!}\cdot\frac{3^n\,\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)}=\sum_{n\geq 0}\frac{1}{6\cdot 2^n}\cdot\frac{\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)\Gamma(n+2)}\tag{2} $$
or, by using Euler's beta function and the reflection formula for the $\Gamma$ function, as
$$ \sum_{n\geq 0}\frac{1}{6\,\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{5}{3}\right)}\cdot\frac{B\left(n+\frac{5}{3},\frac{1}{3}\right)}{2^n}=\frac{\sqrt{3}}{8\pi}\sum_{n\geq 0}\int_{0}^{1}\frac{x^{n+2/3}(1-x)^{-2/3}}{2^n}\,dx \tag{3}$$
from which:
$$ S = \frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{x^{2/3}}{(1-x)^{2/3}(2-x)}\,dx \stackrel{\frac{x}{1-x}\mapsto z}{=}\frac{\sqrt{3}}{4\pi}\int_{0}^{+\infty}\frac{z^{2/3}}{(1+z)(2+z)}\tag{4}$$
and:
$$ S = \frac{\sqrt{3}}{4\pi}\left(-\int_{0}^{+\infty}\frac{z^{2/3}}{z(1+z)}\,dz+2\int_{0}^{+\infty}\frac{z^{2/3}}{z(2+z)}\,dz\right).\tag{5} $$
Euler's beta function then leads to:
$$ S = \color{blue}{\frac{1}{\sqrt[3]{2}}-\frac{1}{2}} \approx 0.2937.\tag{6}$$
