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$G=\{z \in \mathbb{C}: 1<|z|<2\}$. Suppose $\sum_{n=-\infty}^{\infty} \alpha_nz^n$ converges absolutely pointwise on some open neighborhood of $\bar{G}=\{z \in \mathbb{C}: 1 \leq |z| \leq 2\}$. We need to show that $\sum_{n=-\infty}^{\infty} \alpha_nz^n$ converges unifomly on $\bar{G}=\{z \in \mathbb{C}: 1 \leq |z| \leq 2\}$.

Absolute convergence of series gave that $\sum|\alpha_n|$ converges. Now I wish to use the Weierstrass M-test. The uniform convergence of $\sum_{n=1}^{\infty} \frac{\alpha_{-n}}{z^n}$ is done. I am not able to show the uniform convergence of positive power series.

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Any open neighborhood of $\overline {G}$ contains a point with $|z| >2$ so $\sum |\alpha_n| (2+\epsilon)^{n} <\infty$ for some $\epsilon >0$. Now you can apply M-test.

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  • $\begingroup$ It is some open neighborhood in $\bar{G}$, why does it contain a point with $|z|>2$? $\endgroup$
    – Arindam
    Aug 6, 2018 at 23:37
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    $\begingroup$ Let $U$ be a neigborhood of $\overline {G}$. $z=2$ is in $\overline {G}\subset U$ and $U$ is open. Hence there exists $r>0$ such that $|w-2|<r$ implies $w \in U$. In particular $2+\frac r 2 \in U$. $\endgroup$ Aug 6, 2018 at 23:40
  • $\begingroup$ Oh, I was considering an open set inside $\bar{G}$. Thank you. $\endgroup$
    – Arindam
    Aug 6, 2018 at 23:44

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