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I'm trying to find out for which $|z|=1$ the series $$\sum_{n=1}^\infty{}\frac{z^n}{n}$$ converges. It diverges for $z=1$ (harmonic series) and converges for $z=-1$ (alternating harmonic series). I now know that this is the Taylor series for $-log(1-z)$ which doesn't really help me (although it was interesting to find out)?

I found proofs that the series converges for all $|z|=1$ except for this single $z=1$, but we're asked to proof convergence without using test methods we didn't learn (Dirichlet's test, Abel's test) and therefore are not allowed to use. Have you got any hints on how to get to the wanted result more "heuristically"?

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  • $\begingroup$ Is it unbelievable that you have the good idea of multiplying with $1-z$ yourself? $\endgroup$ – Daniel Fischer Nov 28 '14 at 21:38
  • $\begingroup$ You mean $\sum_{n=1}^\infty{}\frac{z^n}{n}=-ln(1-z)$ multiplied with $1-z$? I don't really see how that would help me, but maybe I just can't see the wood for the trees and should go to bed for now. $\endgroup$ – MarocJ Nov 28 '14 at 22:04
  • $\begingroup$ By the way, are you sure that you shall consider complex $z$? If the exercise concerns only real $z$, the case is simple. If you are asked to prove convergence on the entire unit circle except $1$ without Dirichlet's etc. test, I don't really see how you could do that without effectively proving a special [admittedly easier than the general] case of such a test. $\endgroup$ – Daniel Fischer Nov 28 '14 at 22:05
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    $\begingroup$ Not just the difference between consecutive partial sums. You need to show that for every fixed $z$ (on the unit circle except $1$) the sequence $(s_m(z))_{m\in\mathbb{N}}$ of partial sums is a Cauchy sequence. It's easy to show that $\bigl((1-z)s_m(z)\bigr)_{m\in\mathbb{N}}$ is a Cauchy sequence. $\endgroup$ – Daniel Fischer Nov 28 '14 at 22:30
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    $\begingroup$ I'm sure you can figure out when you can just do that if you think about it for a while. $\endgroup$ – Daniel Fischer Nov 28 '14 at 22:49
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Root criteria gives you: $$\sqrt[n]{\lvert \frac{z^n}{n}\rvert} = t $$ with $ \lim\limits_{n \to \infty} \sqrt[n]n = 1$ you get $\lvert z \rvert = t$

Root criteria now says if $t \lt 1$, the series converges and for $t \gt 1$ the series diverges. For $t=1$ the root criteria does not apply, but for $t=1$ only the cases $z=1$ and $z=-1$ exist for which you already know what happens

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  • $\begingroup$ The OP considers complex $z$, where $\lvert z\rvert = 1$ offers far more possibilities. $\endgroup$ – Daniel Fischer Nov 28 '14 at 21:53

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