So my question is $$ \sum_{n=0}^{\infty}\ \left(\frac{2}{z-i}\right)^n$$ this series converges for all z lying inside the disk of radius $2$ centered at i. My doubt here is, what happens at $|z-i|=2$. Is it true that it diverges for every $\theta$ since this series
$$ \sum_{n=1}^{\infty}\ e^{-i\theta n}$$ can be derived if we take $z=i+2e^{i\theta}$ and does not converge for any theta i.e. no $z$ such that $|z-i|=2$.
Yes, you are right. There is no need to mention $e^{-in\theta}$ though. If $z=i+2w$, with $|w|=1$, then you get the series$$\sum_{n=1}^\infty\left(\frac1w\right)^n=\sum_{n=1}^\infty\frac1{w^n},$$which diverges, since you don't have $\lim_{n\to\infty}\frac1{w^n}=0$.