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This question already has an answer here:

Consider the following complex power series $$\sum_{n \geq 1} \frac{z^n}{n} \,\,\,\,\,\,\, z \in \mathbb{C}$$

It surely converges conditionally for $z=-1$ (for alternating series test) and for $z=1$ it diverges (it is the harmonic series).

My question is: how can one show that the power series converges conditionally for any $z \in \mathbb{C}$ such that $|z|=1$ (except for $z=1$)?

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marked as duplicate by Martin R, Daniel Fischer complex-analysis Dec 12 '16 at 19:30

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  • $\begingroup$ The complex version of Abel's test will work for this purpose. $\endgroup$ – Semiclassical Dec 12 '16 at 14:41
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We can use the Dirichlet's test to prove the desired result: with $a_n=\frac1n$ and $b_n=e^{in\theta}$ we verify that

  • $(a_n)$ is decreasing to 0
  • $\sum_{n=1}^N b_n$ is bounded

so the series $\sum a_nb_n$ is convergent.

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