Set of pointwise convergence for power series and their derivatives

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!

• Consider $f_n (z) = \frac{1}{{n^2 }}z^n$.
– Gary
Sep 15 '20 at 7:36

Example: $$\sum_{n=0}^{\infty}\frac{1}{n^2}z^n$$ is convergent for each $$z$$ with $$|z| \le 1$$,
$$\sum_{n=1}^{\infty}\frac{1}{n}z^{n-1}$$ is divergent in $$z=1.$$
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 $$|z|=1$$.