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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.2012-06-20
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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.2012-06-20
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    Yes ,you are right.I have to think over it more.2012-06-20
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    @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 zero2012-06-20
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    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.2012-06-20
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    I have to use a_n converges to 0 .@ Patato ,you are absolutely right what I said earlier is not 100% correct.2012-06-20
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    Here's an idea: Without loss of generality, assume the pole is at 1. Then there is some residue at the pole, and $$f(z)-\frac{A}{z-1}$$ is holomorphic at 1 for some constant A. Consider the power series coefficients of this function in the unit disk. They are bounded away from zero (eventually, as $a_n\rightarrow 0$), so the function can't converge at 1.2012-06-20
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    Pole at 1 need not be simple but then I think same argument will do the job.Am I right?2012-06-20
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    I think so. The coefficients of the power series for $1/(z-1)^n$ get even larger (differentiate the series for $1/(z-1)$) so I don't think you should have any problem completing the argument.2012-06-20
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    Yes,Thanks for your help.2012-06-20
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    No problem, glad to help.2012-06-20

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