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Let $f(z)=\sum_{n=0}^{+\infty} a_n z^n$ has only one pole of order $1$ on the Convergence circumference. Prove that $\sum_{n=0}^{+\infty} a_n z^n$ is divergent on the Convergence circumference.

Let $\zeta$ be the only pole, argue by contradicition, suppose $\sum_{n=0}^{+\infty} a_n z^n$ is convergent at $z_0$ on the Convergence circumference. Then what? Can``$\lim_{z\to \zeta}(z-\zeta)\sum_{n=0}^{+\infty} a_n z^n$ is finite''be used? Thank you.

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  • $\begingroup$ Do you mean "exactly one" pole on covergence circumference? $\endgroup$
    – psl2Z
    Commented May 7 at 11:37
  • $\begingroup$ @psl2Z Yes. Indeed $\endgroup$
    – xldd
    Commented May 7 at 11:58
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    $\begingroup$ Assume by a rotation and dilation that $z=1$;and then if by contradiction $\sum a_nw^n$ converges for some $|w|=1$ it follows $a_n\to0$ but then by Abel theorem we have $(1-z)\sum a_nz^n$ converges to $0$ when $z\to1$ since if you write that as $\sum b_nz^n$ it's immediate that the partial sums of $b_n$ are $a_n$; that is of course a contradiction to simple pole at $1$ (and more generally to any kind of pole at $1$) $\endgroup$
    – Conrad
    Commented May 7 at 12:22

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