Convergence radius with series

Question

Does there exist a complex numbers series $a_n$ , $b_n$ (where n is a natural number) that $\sum_{n=1}^∞ a_nz^n$ $\sum_{n=1}^∞ b_nz^n$ convergence radius would be $1$ and $\sum_{n=1}^∞ (a_n+b_n)z^n$ convergence radius would be $3$ ? Any ideas how to show it?

Answer 1

Yes. Take $a_n=1$ and take$$b_n=\begin{cases}-1&\text{ if $n$ is odd}\\-1+\dfrac1{9^n}&\text{if $n$ is even.}\end{cases}$$

Answer 2

Take the problem upside down:

$a_n=1$ for all $n$ leads to $\sum z^n$ which has RCV $1$.

$c_n=\frac{1}{3^n}$ gives $\sum \left(\frac{z}{3}\right)^n$ which has RCV $3$.

$b_n=c_n-a_n$ will automatically meet the requirements.