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Find the radius of convergence of this series and study what happens in the border. $\sum_{n=1}^{\infty}\frac{z^n}{n}$ ($z\in \Bbb{C}$)

I easily found that the radius of convergence is $\rho =1$, therefore the series doesn't converge absolutely for $|z|=\rho=1$ , since $\sum|\frac{z^n}{n}|$ diverges in this case.

Therefore I want to use a convergence criteria, Dedekind or Dirichlet, but my problem is that the partial sums of $z_n =z^n$ with $|z|=1$ are not bounded.

Any hint?

Martin Sleziak
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Joaquin Liniado
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2 Answers2

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In fact, the partial sums are bounded. Note that for $z$ with $|z| = 1$ and $z \neq 1$, $$ \left|\sum_{k=0}^n z^n\right| = \left|\frac{1 - z^{n+1}}{1 - z}\right| = \frac{|1 - z^{n+1}|}{|1 - z|} \leq \frac{2}{|1-z|} $$ The Dirichlet criterion applies.

Ben Grossmann
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  • Is the first $\le$ not actually an equal sign? And can you explain how you get the last estimate $\frac{2}{|1-z|}$? – mdcq Dec 06 '17 at 13:03
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    @philmcole well spotted. And I used the triangle inequality: $|1 - z^{n + 1}| \leq |1| + |-z^{n+1}| = 2$ – Ben Grossmann Dec 06 '17 at 13:14
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Remember that the radius of convergence is defined as the supremum of convergent series, so there is no need to look on the boundary point itself, if you can get away with a limiting argument. In this case, since it doesn't converge at $z=1$ we know $R\le1$, and conversely we can consider an arbitrary $z$ with $|z|<1$; then $$\sum_{n=1}^\infty\frac{|z|^n}n\le\sum_{n=0}^\infty|z|^n=\frac1{1-|z|}.$$

Mario Carneiro
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  • "Remember that the radius of convergence is defined as the supremum of convergent series" Not sure what you mean. – zhw. Oct 05 '15 at 20:20
  • @zhw. The radius of convergence of $\sum_na_nz^n$ is defined as $R=\sup{z\in\Bbb R\mid\sum_na_nz^n\mbox{ converges}}$. – Mario Carneiro Oct 05 '15 at 20:22
  • Never seen it defined that way. But it's equivalent to the definition I know. – zhw. Oct 05 '15 at 20:24