The problem comes from the Exercise 19 of Stein's Complex Analysis. It is trivial when $z=1$, but I cannot come up with a solution for other cases that $|z|=1$.
My effort is to compute the $\sum^{N}_{n=1} n z^n$, and prove that for any $M \in \mathbb{C}$, there exits some $N$, s.t. $|\sum^{N}_{n=1} n z^n|>|M|$.
Note that when $z\neq 1$, we have $$ |\sum^{N}_{n=1} n z^n| = \frac{|z-(1+N)z^{N+1}+Nz^{N+2}|}{|1-z|^2} $$ But I could not make it out by Triangular inequality.
Thanks in advance
Your approach complicates the problem a lot. I think you should abandon it. Just recall for a series $\sum_{n\ge 0} a_n$ to be convergent, the sequence $(a_n)_{n\ge 0}$ must tend to $0$.
The latter is clearly not true in your example, implying divergence.
For the case that you absolutely want to make your approach work, you can do it like this:
$$|z-(1+N)z^{N+1}+Nz^{N+2}| \ge |-Nz^{N+1}+ Nz^{N+2}| - | z -z^{N+1}|$$ by inverse triangle inequality. Then note this equals, as $|z^{N+1}|=1$, $$N|-1+ z| - | z -z^{N+1}|$$ Finally bound $| z -z^{N+1}| \le 2$ for a lower bound of $$ \ge N|-1+ z| -2.$$ As $|-1+z|$ is non-zero and independent of $N$ this is fine.