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I am confronted with the following problem:

The radius of convergence of a function $f(z)=\sum c_n(z-z_0)^n$ is $R$, and the function has a pole at some $w_0$, with $|w_0-z_0|=R$. Why does, for any other $w$ on the circle with radius $R$, the series not converge absolutely?

I tried to use the MVT, but that does not work out. Can anyone give me a sketch of the proof?

best,

MP

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2 Answers 2

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By assumption $\sum c_n(w_0-z_0)^n$ divergent. This implies that $\sum |c_n||w_0-z_0|^n=\sum |c_n|R^n$ is also divergent.

Choose any other $w$ such that $|w-z_0|=R$. Then we have $\sum|c_n||w-z_0|^n=\sum |c_n|R^n$ which we already known diverge.

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Caution: it's not so obvious that $\sum c_n (w_0 - z_0)^n$ diverges. You need Abel's theorem, I think –  Robert Israel Nov 11 '11 at 0:31
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If the series converges absolutely, $|f(z)| \le \sum_{n=0}^\infty |c_n| R^n$ on $\{z: |z - z_0| < R\}$. But if $w_0$ is a pole, $|f(z)| \to \infty$ as $z \to w_0$.

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