# radius of convergence of a complex power series

Can you tell me what you think about my solution to this problem? In case it's wrong, or needs changes, just something like "try looking at this", "consider that"... a hint is enough, please no complete answers.

Let $R_1$ be the radius of convergence of $\sum a_n z^n$, and $R_2$ the radius of convergence of $\sum b_n z^n$. What can you say about the convergence radius of $\sum (a_n \pm b_n)z^n$?

My work so far:

Assume that $R_1<R_2$.

If $z<R_1$, $\sum a_n z^n$ and $\sum b_n z^n$ converge so $\sum (a_n \pm b_n)z^n$ will also converge.

If $R_1 < z < R_2$ then $\sum a_n z^n$ will diverge and $\sum b_n z^n$ will converge therefore $\sum (a_n \pm b_n)z^n$ will diverge.

If $R_1 < R_2 < z$, both $\sum a_n z^n$ and $\sum b_n z^n$ diverge, so at first I was unsure about what happened in this case, but since all power series have a convergence radius $R$ such that they converge for $z<R$ and diverge for $R<z$, $R_1$ must be the convergence radius for $\sum (a_n \pm b_n)z^n$.

• Suppose $b_n=-a_n$ for every $n\in\{0,1,2,3,\ldots\}$. Then $\sum_n (a_n+b_n)z^n$ has an infinite radius of convergence although $\sum_n a_n z^n$ and $\sum_n b_n z^n$ may have a finite radius of convergence. ${}\qquad{}$ Mar 10, 2015 at 0:00
• @MichaelHardy indeed. For clarity, are you telling me to be aware of this particular case or to reconsider my "solution" because of this case? Mar 10, 2015 at 0:08
• What do you think is the answer here? You only need to address where you know for sure that $\sum_n (a_n+b_n)$ converges. As Michael points out above, the radius of convergence may get larger... Mar 10, 2015 at 0:16
• I was merely mentioning a particular case. Maybe I'll comment on your arguments later. Mar 10, 2015 at 0:18

If the two original series have radii of convergence that are DIFFERENT, then the radius of convergence for $\sum_n (a_n + b_n)z^n$ will be the minimum of the two radii of convergence. If both series have radii of convergence that are the SAME, then the radius of convergence for $\sum_n (a_n + b_n)z^n$ is at least as large but could be larger, due to some cancellations when you add $a_n + b_n$. Consider for example $a_n = (1/2)^n + 1/n$, $b_n = (1/2)^n - 1/n$.