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In complex analysis the definition of radius of convergence is as far as I know given by $$ R = \mathrm{sup} \left\{ |z| : \sum_{n=0}^{\infty} |a_n z^n| \;\text{converges} \right\} $$ So this definition looks at absolute convergence of the power series.

In general we know that a convergent series does not have to be absolutely convergent.

However, I think it can be shown (I can post my proof of this if necessary) that if we define radius of convergence as $$ \tilde R = \mathrm{sup} \left\{ |z| : \sum_{n=0}^{\infty} a_n z^n \;\text{converges} \right\} $$ then $R = \tilde R$.

My question is why is it the case for (complex) power series that for radius of convergence it does not seem to matter if you use absolutely convergent series or convergent series in the definition of radius of convergence?

Edit: Reason I am asking is that in complex analysis we often speak / use absolutely convergent power series. But given the two equivalent definitions, we can just as well speak of convergent power series. The important point is the radius of convergence, right?

Frido
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  • The definitions are equivalent, see for example https://math.stackexchange.com/a/3273162/42969 – Martin R Jun 18 '23 at 08:18
  • @MartinR The answer you linked to only shows / speaks about absolutely convergent series. What I am puzzled by is that im the definition of radius of convergence it doesn't seem to matter if you take the sup over conditionally convergent or absolutely convergent power series. – Frido Jun 18 '23 at 08:22
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    We have \begin{align} \cal R_1&:=\Bigl{|z|:\sum |a_nz^n|\text{ converges}\Bigr}\ \subseteq\cal R_2&:=\Bigl{|z|:\sum a_nz^n\text{ converges}\Bigr}\ \subseteq\cal R_3&:=\Bigl{|z|:(a_nz^n)\text{ converges to $0$}\Bigr}\ \subseteq\cal R_4&:=\Bigl{|z|:(a_nz^n)\text{ is bounded}\Bigr}, \end{align} so $R_1\le R_2\le R_3\le R_4$, where $R_i:=\sup{\cal R_i}$. Now, for all $|z|<|z'|<R_4$, $$\sum|a_nz^n|=\sum\left|a_n{z'}^n\right|{\left|\frac z{z'}\right|}^n \le C\sum{\left|\frac z{z'}\right|}^n<\infty,$$ giving $R_4\le R_1$ and $R_1=R_2=R_3=R_4$. – nejimban Jun 18 '23 at 08:22
  • @nejimban Nice. If you could post your comment as an answer I'll accept it as answer. – Frido Jun 18 '23 at 08:27
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    Does this answer your question? https://math.stackexchange.com/q/659886/42969 – Martin R Jun 18 '23 at 08:29
  • @MartinR Yes, thank you. nejimban's comment I think is also an elegant proof of it. – Frido Jun 18 '23 at 08:33

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