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I have to find the radius of convergence of the power series $\sum_{i=0}^n z^n$ And then discuss the behavior at the edge of the convergence disk.

At the moment I just find that the series converge at the disk of radius 1. So I'm trying to discuss what happen at the edge of the disk.

I consider that $z=(cost,sint)$ so $z^n=(cos(nt),sin(nt)), 0 \le t \le 2\pi$.

Then I conclude that $\sum_{i=0}^n z^n$ converge if $\sum_{i=0}^n cos(nt)$ and $\sum_{i=0}^n sin(nt)$ converge.

I started my discussion as follows:

if $t=0$ the series not converge because $cos(n0) = cos (0) = 1$ so $\sum_{i=0}^n cos(nt)$ = $\sum_{i=0}^n 1$ not converge

if $t=\pi$ the series not converge because $cos(n\pi) = -1$ so $\sum_{i=0}^n cos(nt)$ = $\sum_{i=0}^n -1$ not converge

if $t=\pi/2$ the series not converge because $sin(n\pi/2) = 1$ so $\sum_{i=0}^n sin(nt)$ = $\sum_{i=0}^n 1$ not converge

if $t=3\pi/2$ the series not converge because $cos(n3\pi/2) = -1$ so $\sum_{i=0}^n sin(nt)$ = $\sum_{i=0}^n -1$ not converge

Now my question is how can I prove that the series $\sum_{i=0}^n cos(nt)$ and $\sum_{i=0}^n sin(nt)$ not converge?

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    You need to rewrite all of those series; they make no sense as it stands. – 2017-02-16
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    Why do I have to rewrite these series? – 2017-02-16
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    You don't see why $\sum_{i=0}^n cos(nt)$ is wrong? – 2017-02-16

1 Answers 1

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If $|z|=1$ then $|z^n|=1$ and $z^n$ does not converge to $0$. Hence, the series $\sum_{n=0}^\infty z^n$ does not converge.

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    Thank you. So it is not necessary to consider $z=(cos t,sin t)$ – 2017-02-16