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Consider the power series $\sum a_n z^n$.Given that $a_n$ converges to $0$, prove that $f(z)$ cannot have pole on the unit circle, where $f(z)$ is the function represented by the power series in the question.

EDIT

I have thought an answer for it. Since $a_n$ converges to $0$, we can write $\lvert a_n \rvert <1$ for all $n >N_0$. From here, we can say radius of convergence of the power series is bigger than or equal to $1$. If the radius of convergence is bigger than $1$, the series converges on the unit circle. If it is equal to $1$, then points on the unit circle cannot be an isolated singularity. But I am not sure of my answer.

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Shraddha, Welcome to math.SE. since you are a new user, we wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, people are also much happier to help those who demonstrate that they've tried the problem themselves first. –  user17762 Jun 20 '12 at 9:13
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@ShraddhaSrivastava Depending on how you interpret the question, I don't buy your reasoning for the case when the radius of convergence is equal to 1. A function can have radius of converge 1 around the origin and have an isolated singularity because of analytic continuation. Consider $1/(1-x)$ for example. –  Potato Jun 20 '12 at 9:41
    
Yes ,you are right.I have to think over it more. –  Shraddha Srivastava Jun 20 '12 at 9:44
    
@Potato, the power series in this case is$$\frac{1}{1-z}=\sum_{k=0}^\infty z^n\,\,,\,|z|<1$$ and here $\,\{a_n=1\}\,$ doesn't converge to zero –  DonAntonio Jun 20 '12 at 9:47
    
That is exactly my point. His claim is that any function whose radius of convergence is exactly 1 cannot have isolated singularities on the unit circle. –  Potato Jun 20 '12 at 9:48
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1 Answer 1

A power series whose coefficients tend to $0$, and whose radius of convergence is therefore at least $1$, can define a function that has isolated sigularities on the unit circle; only those isolated singularities cannot be poles. So your reasoning is not correct, and this claim needs to be changed.

Take for instance the power series with $a_0=0$ and $a_n=\frac1n$ for $n>0$. This defines the function $f: z\mapsto\ln(\frac1{1-z})$ which has an isolated singularity at $1$.

So instead of singularities, you should be thinking about what having a pole means.

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Yes,It can have isolated singularity on the unit circle but that singularity can not be pole.I was not precise(Infact I was wrong) –  Shraddha Srivastava Jun 20 '12 at 10:16
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