A series is given as follows 16+56⋅12+5⋅86⋅12⋅18+5⋅8⋅116⋅12⋅18⋅24+…

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

For any $n\in {\mathbb{N}}^{+}$ we have
$\prod _{k=1}^n(3k+2)=\frac{3^n\mathrm{\Gamma }(n+\frac{5}{3})}{\mathrm{\Gamma }(\frac{5}{3})}$
and the value of the RHS at n=0 equals 1. It follows that the whole series can be represented as
$S=\sum _{n\ge 0}\frac{1}{6^{n+1}(n+1)!}\cdot \frac{3^n\mathrm{\Gamma }(n+\frac{5}{3})}{\mathrm{\Gamma }(\frac{5}{3})}$
$=\sum _{n\ge 0}\frac{1}{6\cdot 2^n}\cdot \frac{\mathrm{\Gamma }(n+\frac{5}{3})}{\mathrm{\Gamma }(\frac{5}{3})\mathrm{\Gamma }(n+2)}$
or, by using Euler's beta function and the reflection formula for the $\mathrm{\Gamma }$ function, as
$\sum _{n\ge 0}\frac{1}{6\mathrm{\Gamma }(\frac{1}{3})\mathrm{\Gamma }(\frac{5}{3})}\cdot \frac{B(n+\frac{5}{3},\frac{1}{3})}{2^n}=\frac{\sqrt{3}}{8\pi }\sum _{n\ge 0}{\int }_0^1\frac{x^{n+2/3}(1-x{)}^{-2/3}}{2^n}dx$
from which:
$S=\frac{\sqrt{3}}{4\pi }{\int }_0^1\frac{x^{2/3}}{(1-x{)}^{2/3}(2-x)}dx=\frac{\sqrt{3}}{4\pi }{\int }_0^{+\mathrm{\infty }}\frac{z^{2/3}}{(1+z)(2+z)}$
and:
$S=\frac{\sqrt{3}}{4\pi }(-{\int }_0^{+\mathrm{\infty }}\frac{z^{2/3}}{z(1+z)}dz+2{\int }_0^{+\mathrm{\infty }}\frac{z^{2/3}}{z(2+z)}dz)$
Euler's beta function then leads to:
${\displaystyle S=\frac{1}{\sqrt[3]{2}}-\frac{1}{2}\approx 0.2937}$