# Convergence radius with series [closed]

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?

## closed as off-topic by Martin R, Namaste, Saad, JonMark Perry, Chris CusterApr 6 '18 at 3:02

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