Critical series

Let $S$ be the sum given. Consider the Taylor series centered around $x=0$ of $\sqrt[3]{1+x}$ :

$\begin{aligned} \sqrt[3]{1+x} & = 1 + \frac x3 - \frac {2x^2}{2!3^2} + \frac {2\cdot 5 x^3}{3!3^3} - \frac {2\cdot 5 \cdot 8 x^4}{4!3^4} + \cdots & \small \color{#3D99F6} \text{Differentiate both side w.r.t. }x \\ \frac 13 (1+x)^{-\frac 23} & = \frac 13 - \frac {2x}{1!3^2} + \frac {2\cdot 5 x^2}{2!3^3} - \frac {2\cdot 5 \cdot 8 x^3}{3!3^4} + \cdots & \small \color{#3D99F6} \text{Multiply both side by }3 \\ (1+x)^{-\frac 23} & = 1 - \frac {2x}{1!3} + \frac {2\cdot 5 x^2}{2!3^2} - \frac {2\cdot 5 \cdot 8 x^3}{3!3^3} + \cdots & \small \color{#3D99F6} \text{Put } x = - \frac 13 \\ \left(1-\frac 13\right)^{-\frac 23} & = 1 +\frac {2}{1!3^2} + \frac {2\cdot 5}{2!3^4} + \frac {2\cdot 5 \cdot 8}{3!3^6} + \cdots \end{aligned}$

$\begin{aligned} \implies S & = \left(1-\frac 13\right)^{-\frac 23} = \left(\frac 32\right)^\frac 23 = \boxed{\sqrt[3]{\dfrac 94}} \end{aligned}$

n=-2/3

x=-1/3

(1-x)^n =(9/4)^(1/3)