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Consider the following complex power series :$\sum_{n=1}^\infty\frac{z^n}{n}$ The radius of convergence of this series is $1$ and the series is divergent for $z=1$. I want to know what are the values of $z\in C:=\lbrace z\in\mathbb{C}: |z|=1\rbrace$, the circle of convergence, for which the given series converges.

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    To find any value you may rewrite it as $\ -\ln(1-z)\ $ (for $z\not =1$)2012-09-10

2 Answers 2

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HINT: Look at Dirichlet test. In your case, choose $a_n = \dfrac1n$ and $b_n = z^n$.

From the Dirichlet test, you will get that the series converges everywhere on the boundary of the unit disc except at $z=1$.

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    Although it ultimately amounts to the same, I'd suggest to consider $(1-z)f(z)$ where $f(z)$ is the given series.2012-09-10
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Hint: This is a classical example for Abel's Test.

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    According to [this statement of Abel's Test](http://en.wikipedia.org/wiki/Abel%27s_test), it doesn't apply, since $\sum\limits_{n=1}^\infty z^n$ doesn't converge. Would you explain what you mean by Abel's Test?2012-09-14