This is very important remark, based on my reevaluation of this problem

1. $|z-z_0|=R$ is a circumference, not just two points. Those who found correctly $R$ but checked $z=-R$ and $z=R$ got only half-mark (and rightfully so!)

2. At $|z-z_0|=R$ neither root, nor ratio criteria work.

This would make convergence of $\sum_{n=1}^\infty n^{-p} z^n$ more difficult to check than in the real case for $0<p\le 1$. However this series diverges for all $z\colon |z|=1$ for $p=0$ and converges for all $z\colon |z|=1$ for $p>1$. Criteria, respectively: *the term does not tend to $0$* and *if the series converges absolutely, it converges* where *absolute convergence* means convergence of the series made of absolute values.

3. If $R=0$ it means that the series converges iff $z=z_0$, or diverges everywhere except $z=z_0$. Those who claim that "diverges everywhere" without making an exception for $z=z_0$ got half-mark (and rightfully so)