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Here's Theorem 3.44 in the book Principles of Mathematical Analysis by Walter Rudin, third edition:

Suppose the radius of convergence of $\sum c_n z^n $ is $1$, and suppose $c_0 \geq c_1 \geq c_2 \geq \cdots$, $\lim_{n\to\infty} c_n = 0$. Then $\sum c_n z^n$ converges at every point on the circle $\vert z \vert = 1$, except possibly at $z = 1$.

Now I have the following couple of questions.

(1) Can we replace the second condition in the above theorem by the following? Suppose $\vert c_0 \vert \geq \vert c_1 \vert \geq \vert c_2 \vert \geq \cdots$. Does the conclusion of the above theorem still hold?

Let $\{c_n\}_{n\in\mathbb{N}}$ be a sequence of complex numbers; suppose the power series $\sum c_n z^n$ has radius of convergence equal to $R$, where $R$ can either be a non-negative real number or $+\infty$. Then we can we say $R$ is also the radius of convergence of the power series $\sum \vert c_n \vert z^n$. Am I right?

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The two series $$ f(z)=\sum \frac{z^n}{n},\quad g(z)=\sum (-1)^n\frac{z^n}{n} $$ are a counterexample to your conjecture, because $f(z)$ converges everywhere on the unit disk except at $z=1$ while $g(z)$ converges everywhere on the unit disk except at $z=-1$ (note, indeed, that $f(-z)=g(z)$).

The answer to your second question is affirmative, as one can readily see from Cauchy-Hadamard's characterization of the radius of convergence.

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  • $\begingroup$ It would be interesting to know if the condition $|c_1|>|c_2|>\ldots \to 0$ implies that the series converges everywhere on the unit disk except at a single point (which is $1$ if $c_n\ge 0$ for all $n$). $\endgroup$ May 25 '16 at 12:22

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