$\sum \frac{z^k}{k}$ does not converge uniformly on $|z|<1$

Question

$\sum \frac{z^k}{k}$ does not converge uniformly on $|z|<1$

In this post, the answer says that as $\sum\frac{1}{n}$ does not converges, $\lim_{z\to 1^-,z\in\Bbb R}\sum\frac{z^k}{k}= \infty$. Why is this a valid proof?

If I have some series $\sum a_nz^n$ and I'm asked to show the series does not converge uniformly on $|z|<R$, then if I show $\sum a_nR^n$ does not converges, does that show the series $\sum a_nz^n$ does not converge uniformly?

Answer 1

Proving non-uniform convergence in this way follows from a simple fact.

Suppose the partial sums are bounded, that is, there exists a sequence $(M_n)$ of positive real numbers such that $|S_n(z)| = \left| \sum_{k=1}^n f_k(z)\right| \leqslant M_n$ for all $n \in \mathbb{N}$ and for all $z \in D$. If $S_n(z) \to S(z)$ uniformly for $z \in D$, then $S$ must be bounded on $D$.

To prove this, note that by uniform convergence there exists $N \in \mathbb{N}$ such that $|S_N(z) - S(z)| \leqslant 1$ for all $z \in D$. Hence $|S(z)| \leqslant |S_N(z) - S(z)| + |S_N(z)| \leqslant 1+ M_N$ for all $z \in D$.

Here we have $|S_n(z)| = \left|\sum_{k=1}^n \frac{z^k}{k} \right|\leqslant \sum_{k=1}^n \frac{|z|^k}{k} \leqslant \sum_{k=1}^n \frac{1}{k}\leqslant n = M_n$ for all $|z| < 1$. If the series converged uniformly the sum would have to be bounded for all $|z| < 1$ -- which it is clearly not.