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 $\theta = arg (z)$?
Show that $\sum_{n = 1}^\infty \frac {z^n} {n}$ converges when $|z| = 1$ , except when $z =1$
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complex-analysis
power-series
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0Hint: It does not converge at $z=1$. – 2017-02-04
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0It surely doesn't converge for $z=1$. – 2017-02-04
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0I'm not convinced on this... Let $z=1$ and then we have the Harmonic series – 2017-02-04
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0It's convergent when $|z|=1$ with $z\neq 1$. – 2017-02-04
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0Made the changes. When $z = 1$ it's a harmonic series that's known to diverge. – 2017-02-04