I'm stuck at this one. If $z \ne 1$, can I show this by invoking that when $|z| = 1$ then $z = e^ {i \theta}$ where $\theta = arg (z)$?
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Hint: It does not converge at $z=1$. – Simply Beautiful Art Feb 04 '17 at 17:27
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It surely doesn't converge for $z=1$. – Sergei Golovan Feb 04 '17 at 17:27
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I'm not convinced on this... Let $z=1$ and then we have the Harmonic series – Brevan Ellefsen Feb 04 '17 at 17:28
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It's convergent when $|z|=1$ with $z\neq 1$. – Frank Lu Feb 04 '17 at 17:28
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Made the changes. When $z = 1$ it's a harmonic series that's known to diverge. – Harry Evans Feb 04 '17 at 17:29