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I am studying the complex series $$\sum_{n = 1}^{\infty} \frac{z^n}{n}$$ when $|z| = 1$ and $z \ne 1$.

I wrote down from class that in this case the series converges by Dirichlet's test. But I don't see why there exists an $M \in R$ s.t. $\Bigg| \sum_{n = 1}^{N} z^n \Bigg| \le M$ for every positive integer $N$?

Monolite
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2 Answers2

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Well, $\vert\sum_1^N z^n\vert=\vert\frac{z-z^{N+1}}{1-z}\vert\le\frac{2}{\vert 1-z \vert}$ by the triangle inequality and $\vert z\vert=1$. For $z\neq1$, this is finite, and Dirichlet's test applies.

πr8
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  • Thank you a lot for your answer, I have a follow up question: what if we had $\sum_{n = 1}^{\infty} \frac{(-z)^n}{n}$ ? how would I show the existence of $M$ for this series? – Monolite Jan 20 '16 at 14:19
  • If the first series you were summing was $f(z)$, then this new one is just $f(-z)$, so all convergence properties just carry over (though obviously for this new series, it's not convergent at $z=-1$, rather than at $z=1$). – πr8 Jan 20 '16 at 14:21
  • Why do the convergence properties carry over? sorry I am a bit doubtful with complex series since I have only recently been introduced. – Monolite Jan 20 '16 at 14:26
  • If we call $f(z)=\sum_n \frac{z^n}{n}$ and $g(z)=\sum_n \frac{(-z)^n}{n}$, then the series for $g(z)$ is the same as the series for $f(-z)$, and so will be just as convergent (or not) as that series. – πr8 Jan 20 '16 at 14:29
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On the unit circle let $z=e^{2 \pi i t}$. Then, w.l.o.g $$ \left|\sum_{n=1}^Ne^{2\pi i n t}\right|=\left|\frac{e^{2\pi i t}-e^{2\pi i (N+1)t}}{1-e^{2\pi i t}}\right|\leq\frac{2}{|1-e^{2\pi i t}|} $$ for all $N\in\mathbb{N}$. Which then shows that the criteria is met $\forall z \neq 1$ in this circle.