Given a complex power series $\sum_n a_n z^n$ for $z,a_n\in\mathbb{C}$, let $$A:=\{r\geq 0: \exists M\in\mathbb{R}[\forall n\in \mathbb{N} [\, |r^n a_n|<M\ ]] \}$$ $$B := \{ r\geq 0 :\sum_n |r^n a_n| \in \mathbb{R} \}$$
Prove that $\sup A = \sup B$.
Now my colleagues and I are becoming crazy with this one. We know that $B\subset A$, which implies that $\sup B \leq \sup A$, so I am trying to prove that $\sup B < \sup A$ leads to a contradiction, but after many trials, I can't find a reason to imply the conclusion, although I am not able either to find a contre-example to $\sup A = \sup B$, for every time I try I get simply that the radius of convergence of the failed counter-example series is actually the $\max A$, which still confirms the expected result.
On the other hand,
$$\begin{split} &r>\sup B\ &\Rightarrow l :=\limsup_{n\to\infty} |a_n r^n|\neq 0\\ &l \in \mathbb{R} &\Rightarrow \exists M\in\mathbb{R}[\forall n\in\mathbb{N}[\ |a_n r^n| < M\ ]]\\ &&\Rightarrow l \in A\\ &&\Rightarrow \sup A \neq \sup B \end{split}$$
which is a reasoning that I find correct, but I guess it is not, right? I am completely puzzled, and I've been thinking in circles for hours. Please, is $\sup A = \sup B$? Give a proof, or at least lead me to it. Thank you in advance.
Clearly $B\subseteq A$ and hence $\sup B\leq\sup A$. We go to show that $\sup A\leq\sup B$. Prove by contradiction. Suppose the contrary that $\sup A>\sup B$. Choose $r_{1}$ such that $\sup B<r_{1}<\sup A$. Since $r_{1}$ is not an upper bound of $A$, there exists $r\in A$ such that $r_{1}<r$. Choose $M>0$ such that $|r^{n}a_{n}|<M$ for all $n$. Observe that \begin{eqnarray*} & & \sum_{n=0}^{\infty}\left|r_{1}^{n}a_{n}\right|\\ & = & \sum_{n=0}^{\infty}\left|\left(\frac{r_{1}}{r}\right)^{n}r^{n}a_{n}\right|\\ & \leq & \sum_{n=0}^{\infty}M\left(\frac{r_{1}}{r}\right)^n\\ & < & \infty \end{eqnarray*} because $0<\frac{r_{1}}{r}<1$. This shows that $r_{1}\in B$, which is a contradiction.