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 $\sum (a_n \pm b_n)z^n$.

  • 2
    $\begingroup$ 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. ${}\qquad{}$ $\endgroup$ Mar 10, 2015 at 0:00
  • $\begingroup$ @MichaelHardy indeed. For clarity, are you telling me to be aware of this particular case or to reconsider my "solution" because of this case? $\endgroup$
    – notacat
    Mar 10, 2015 at 0:08
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    $\begingroup$ 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... $\endgroup$
    – copper.hat
    Mar 10, 2015 at 0:16
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    $\begingroup$ I was merely mentioning a particular case. Maybe I'll comment on your arguments later. $\endgroup$ Mar 10, 2015 at 0:18

1 Answer 1


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 $a_n = (1/2)^n + 1/n$, $b_n = (1/2)^n - 1/n$.


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