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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?

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Did you mean $|z| < 1$?? Otherwise your question is reaaaaally weird. –  Patrick Da Silva Nov 8 '12 at 7:33

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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: $$ \sum_{k=1}^\infty\frac{z^k}k=-\log(1-z) $$

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Indeed not. Uniform convergence in the closed $(1-\varepsilon)$-disk does not mean uniform convergence over the open $1$-disk. What would be right is this : let $\varepsilon > 0$. If you choose any $\delta > 0$ and wish to find $z_1, z_2$ with $|z_1 - z_2| < \delta$ and $|\sum_{k} z_1^k/k - \sum_k z_2^k/k| \ge \varepsilon$, notice that if you take $z_1 = x$ and $z_2 = (1-\frac {\delta}2)x$ ($x$ is meant to be real here), then $$ \left| \sum_k \frac {x^k}k - \sum_k \frac{\left( 1 - \frac {\delta}2 \right)^k x^k}k \right| = \left| \sum_k \frac{ \left( 1 - (1-\delta/2)^k \right) x^k}k \right| \to \infty $$ as $x \nearrow 1$, so in particular we can find an $x$ so that the latter is $\ge \varepsilon$. Therefore you cannot hope to bound the difference no matter how close $z_1$ and $z_2$ get.

Hope that helps,

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