How can I prove that $\sup(A) = \sup(B)$

Question

How can I prove that $\sup(A) = \sup(B)$

where $A=\{r\in\mathbb{R}^+ : \sum |a_n|r^n \text{ converges}\}$

and $B=\{r\in\mathbb{R}^+: \sum a_n r^n \text{ converges}\}$

I have proved the first part where $\sup(B) ≥ \sup(A)$.

Can I get help to complete the proof? Thanks in advance.

Answer 1

Note $A$ and $B$ must be at least $0$. If $ r > 0$ and $\sum a_n r^n$ converges, then the $|a_n r^n|$ are bounded above by some constant $C$. So, if $0 \leq s < r$, then $$\sum |a_n|s^n = \sum |a_nr^n|\frac{s^n}{r^n} \leq C\sum \left(\frac{s}{r}\right)^n < \infty $$