It's easy to see that the series $\sum \frac{z^k}{k}$ converges locally for |z|<1, by comparison with $\sum z^k$. I'm trying to show why it doesn't converge uniformly. Would it be correct to say uniform convergence for |z|<=1-$\epsilon$ implies uniform convergence for |z| = 1?
[Math] Show that $\sum \frac{z^k}{k}$ does not converge uniformly for |z|<1
complex numberscomplex-analysis
Best Answer
Hint 1: Note that the Harmonic Series diverges, $$ \sum_{k=1}^\infty\frac1k=\infty $$ and that the terms of the series $$ \sum_{k=1}^\infty\frac{z^k}k $$ increase monotonically to those of the harmonic series as $z\to1^-$ along $\mathbb{R}$. Thus, $$ \lim_{\substack{z\to1^-\\z\in\mathbb{R}}}\sum_{k=1}^\infty\frac{z^k}k=\infty $$
Hint 2: $$ \sum_{k=1}^\infty\frac{z^k}k=-\log(1-z) $$