This sequence converges to 0 because the absolute value converges to 0. Note ${\sqrt{\frac{2}{3}}} < 1$, so powers of ${\sqrt{\frac{2}{3}}}^n$
Proof:
The Arg of $\frac{(1+i)}{\sqrt{3}}$ is $\frac{\pi}{4}$. That can be verified as $(1+i)$ "points northeast", and the $\sqrt{3}$ denominator is irrelevant to the Arg.
The magnitude $|\frac{(1+i)}{\sqrt{3}}|$ is ${\sqrt{\frac{2}{3}}}$. That can be verified as $|1+i|$ = $\sqrt2$, and $\frac{\sqrt2}{\sqrt3}$ = ${\sqrt{\frac{2}{3}}}$.
Therefore, $\frac{(1+i)}{\sqrt{3}} = {\sqrt{\frac{2}{3}}}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$
And by DeMoivre's law, $[\frac{(1+i)}{\sqrt{3}}]^n$ = ${\sqrt{\frac{2}{3}}}^n(\cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4})$
Intuitively magnitude of $[\frac{(1+i)}{\sqrt{3}}]^n$ can only spiral down counterclockwise as $n$ increases, and eventually approaches 0.
