Answer
$\Sigma_{n=0}^\infty \dfrac{(-1)^n (2)^{2n+1} (x)^{2n+1}}{3^{2n+1}(2n+1)!}$
Work Step by Step
Since, we know that the Taylor Series for $\sin x$ is defined as:
$ \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!}+...$
Now, $ \sin (\dfrac{2x}{3})=\Sigma_{n=0}^\infty \dfrac{(-1)^n ((\dfrac{2x}{3}))^{2n+1}}{(2n+1)!}=(\dfrac{2x}{3})-\dfrac{(\dfrac{2x}{3})^3}{3!}+\dfrac{(\dfrac{2x}{3})^5}{5!}-\dfrac{(\dfrac{2x}{3})^7}{7!}+...$
or, $=\Sigma_{n=0}^\infty \dfrac{(-1)^n (2)^{2n+1} (x)^{2n+1}}{3^{2n+1}(2n+1)!}$
