Answer
$\Sigma_{n=0}^\infty (-1)^n\dfrac{ (2)^{(2n+1)} (x)^{(2n+1)}}{3^{(2n+1)}(2n+1)!}$
Work Step by Step
Consider the Taylor Series for $\sin x$ as follows:
$ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Then, we have $ \sin (\dfrac{2x}{3})=\Sigma_{n=0}^\infty \dfrac{(-1)^n ((\dfrac{2x}{3}))^{2n+1}}{(2n+1)!}$
$\implies (\dfrac{2x}{3})-\dfrac{(\dfrac{2x}{3})^3}{3!}+\dfrac{(\dfrac{2x}{3})^5}{5!}-\dfrac{(\dfrac{2x}{3})^7}{7!}+...=\Sigma_{n=0}^\infty (-1)^n\dfrac{ (2)^{(2n+1)} (x)^{(2n+1)}}{3^{(2n+1)}(2n+1)!}$
