Let $z \in \mathbb{C} \setminus \{0\}$ and take $z^n = e^{n\operatorname{Log z}}$ where $\operatorname{Log}$ is the principal branch of $\log$. We denote the positive real infinity as $+\infty$ and the complex infinity as $\infty$.
Define $$\operatorname{P.V.} \sum_{n = 0}^{+\infty} \frac{1}{n z^n} = \lim_N \left(\frac{N}{z^{\frac{1}{N}}} + \sum_{n = 1}^N \frac{1}{n z^n}\right)$$ We wish to find out if $$\operatorname{P.V.} \sum_{n = 0}^{+\infty} \frac{1}{n z^n} = \infty$$ In other words, does $$\frac{1}{\frac{N}{z^{\frac{1}{N}}} + \sum_{n = 1}^N \frac{1}{n z^n}} \to 0$$ as $N \to +\infty$?
We know $$\lim_N \sum_{n = 1}^N \frac{1}{n z^n} = -\operatorname{Log}\left(1 - \frac{1}{z}\right)$$ for $|z| > 1$ by Taylor series expansion.
Hence for a fixed $z$, $\left(\sum_{n = 1}^N \frac{1}{n z^n}\right)$ converges to a complex constant and $\left(z^{\frac{1}{N}}\right) \to 1$ as $N \to +\infty$. Thus we can conclude that $$\lim_N \left(\frac{N}{z^{\frac{1}{N}}} + \sum_{n = 1}^N \frac{1}{n z^n}\right) = \infty$$ for $|z| > 1$.
In fact, the same argument works when $|z| = 1$ and $z \neq 1$, because by Determining precisely where $\sum_{n=1}^\infty\frac{z^n}{n}$ converges?, $\left(\sum_{n = 1}^N \frac{1}{n z^n}\right)$ converges to a complex constant when $|z| = 1$ and $z \neq 1$.
For z = 1, $\left(\sum_{n = 1}^N \frac{1}{n z^n}\right)$ is the harmonic series which diverges to $+\infty$ as $N \to +\infty$ and $z^{\frac{1}{N}} = 1$ and hence $$\lim_N \left(\frac{N}{z^{\frac{1}{N}}} + \sum_{n = 1}^N \frac{1}{n z^n}\right) = \infty$$
How about for $|z| < 1$?
