# Uniform convergence of complex series

Been studying some complex series recently, and still quite unsure how to fully show problems such as:

Show that the following series does not converge uniformly on $B(0,1):$

$$\sum_{n=1}^{\infty} \frac{z^n}{n}$$

How do we properly write a proof to show such series does not converge uniformly?

Note that for real $z_n = (1 - 1/n) \in B(0,1)$ as $n \to \infty$
$$\left|\sum_{k=n+1}^{2n}\frac{z_n^k}{k} \right| > \frac{n}{2n}(1- 1/n)^{2n} \to \frac{e^{-2}}{2} \neq 0$$
For uniform convergence it is necessary that for any $\epsilon > 0$ there exists a positive integer $N$ such that for all $m > n > N$ and for all $z \in B(0,1)$ we have
$$\left|\sum_{k=n+1}^{m}\frac{z^k}{k} \right| < \epsilon$$
Assume that it were, then it is uniformly Cauchy, we have for any $\epsilon>0$ some $N$ such that \begin{align*} \left|\sum_{k=n}^{m}\dfrac{z^{k}}{k}\right|<\epsilon \end{align*} for all $z$ with $|z|<1$ and $m\geq n\geq N$. Then letting $z\rightarrow 1^{-}$ we have \begin{align*} \sum_{k=n}^{m}\dfrac{1}{k}\leq\epsilon \end{align*} for all $m\geq n\geq N$, but this means that the sum $\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k}$ is convergent, a contradiction.