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Consider the power series $\sum_{n=0}^{+\infty} f_n(z)=\sum_{n=0}^{+\infty} a_nz^n$.

We define $P=\{z \in \mathbb{C} \mid \sum_{n=0}^{+\infty} f_n(z) \text{ converges}\}$, $P'=\{z \in \mathbb{C} \mid \sum_{n=0}^{+\infty} f_n'(z) \text{ converges}\}$.

Is it always true that $P=P'$?

I know that $\sum_{n=0}^{+\infty} f_n$ and $\sum_{n=0}^{+\infty} f_n'$ have the same radius of convergence, but maybe we can still have that $P \neq P'$.

Namely, maybe we can find $\sum_{n=0}^{+\infty} f_n$ with radius of convergence $R \in (0,+\infty)$ such that $\sum_{n=0}^{+\infty} f_n(R)$ converges, but at the same time $\sum_{n=0}^{+\infty} f_n'(R)$ doesn't converge, and so we have $P \neq P'$.

Thank you!

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    $\begingroup$ Consider $f_n (z) = \frac{1}{{n^2 }}z^n $. $\endgroup$
    – Gary
    Sep 15 '20 at 7:36
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Example: $ \sum_{n=0}^{\infty}\frac{1}{n^2}z^n$ is convergent for each $z$ with $|z| \le 1$,

but

$ \sum_{n=1}^{\infty}\frac{1}{n}z^{n-1}$ is divergent in $z=1.$

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The series $\sum_{n=0}^{\infty}z^n$ has radius of convergence $1$ and diverges at each point on the circle $|z|=1$.

The series $\sum_{n=1}^{\infty}\frac1n z^n$ has radius of convergence $1$ and diverges at $z=1$ but converges for all other $z$ with $|z|=1$.

The first series is the derivative of the second series.

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