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given $\displaystyle{ z \in \mathbb{C}, |z|=1 , z\ne 1 ; \sum_{n\in \mathbb{N}} \frac{z^{n}}{n}}$ I will show that:

i) the partial sum of this series is bounded. , ii) that $\displaystyle{\sum_{n\in \mathbb{N}}\frac{z^{n}}{n}} converges$, iii) $\displaystyle{\sum_{n\in \mathbb{N}}\frac{sin(an)}{n} }$ converges.

I know this is a power series...

i) How can one show the partial sum is bounded?

ii) $\displaystyle{|\sum \frac{z^{n}}{n}|} \le \sum|\frac{z^{n}}{n}| $

iii) write in polar form cos+isin then it follows from ii)

Does anybody see a way? Please do tell.

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    The sum on the right-hand side of ii) doesn't converge for $|z|=1$2011-11-14

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Use summation by parts, also known as Abel's lemma. In the notation of the article in the link, let $a_n=z^n$ and $b_n=1/n$. All you need is an expression for $A_n=z+z^2+\dots z^N$. This is the technique used to prove Dirichlet's criterion for the convergence of series with arbitrary sign.

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    Ah, you've beat me to the punch. I'll just edit in a link to Dirichlet's test.2011-11-14