Let $P(z) = \sum_{n = 0}^\infty c_n z^n$ be a complex power series. Consider the follwing subsets of $\mathbb{R}$

$$ \begin{align} A_1 &:= \{r \geq 0 \,:\, (c_n r^n)_{n \in \mathbb{N}_0} \text{ is a bounded sequence}\} \\ A_2 &:= \{ |z| \geq 0 \,:\, P(z) \text{ converges}\} \end{align} $$

I have two definitions of the radius of convergence $$ \begin{align} R_1 &:= \sup A_1\\ \text{and } R_2 &:= \sup A_2 \end{align} $$ and I want to prove their equivalence, i.e. $R_1 = R_2$.

One way to prove this could be by showing $R_1 \leq R_2$ and $R_1 \geq R_2$.

  1. $R_2 \leq R_1$: This can be shown by proving the inclusion $A_2 \subseteq A_1$. If $z \in \mathbb{C}$ such that $|z| \in A_2$, the convergence of $P(z)$ implies that the terms $c_n |z|^n$ form a null sequence. Hence, $(c_n |z|^n)_{n \in \mathbb{N}_0}$ is bounded.

  2. $R_1 \leq R_2$: For every $|z| < R_1$ there exists $\rho \in A_1$ with $|z| < \rho \leq R_1$. Then by definition of $A_1$ there exists $C \in \mathbb{R}$ such that $|c_n \rho^n| \leq C$ for all $n \in \mathbb{N}_0$ and $$ |c_n z^n| = |c_n \rho^n| \cdot {\underbrace{\left|\frac{z}{\rho}\right|}_{< 1}}^n \leq C q^n $$ with $|q| < 1$. So the series converges by majorization with a geometric series. Hence, for every $|z| < R_1$ we have $|z| \leq R_2$. Using the lemma mentioned by Dominik, this implies $R_1 \leq R_2$.

    • So far, is this proof correct?
    • Do you maybe know a better way to prove this?

1 Answer 1


The first proof needs some improvement. Note that $|z| \le R_2$ does not imply the convergence of $P(z)$. Only if $|z| < R_2$ you can infer the convergence of the power series [and this needs to be proven!].

For the second part you first need to show that the sequence stays bounded for any $z$ with $|z| < R_1$. This is easy to prove, but not trivial.

You don't need to do anything different if the supremum is actually a maximum. You implicitly used the following lemma:
If for all $a < x$ we also have $a \le y$, then $x \le y$.

  • $\begingroup$ First paragraph of your post: $R_2$ instead of $R_1$? $\endgroup$
    – el_tenedor
    Aug 16, 2015 at 11:44
  • $\begingroup$ Indeed, I will correct it. $\endgroup$
    – Dominik
    Aug 16, 2015 at 11:45
  • $\begingroup$ Thanks for pointing out the lemma. I think the first part of my poof was actually showing the other inequality. I have updated my post, correcting the error and (hopefully) completing the proof. $\endgroup$
    – el_tenedor
    Aug 16, 2015 at 12:17
  • $\begingroup$ There is only one minor mistake. While it is true that in the second part, $c_n\rho^n$ is bounded for every $\rho < R_1$, you didn't prove it. The definition of $R_1$ only says that for every $z$ with $|z| < R_1$ there exists a $\rho$ with $|z| < \rho < R_1$ so that $c_n\rho^n$ is bounded. However, in your proof it is sufficient to consider this specific $\rho$. $\endgroup$
    – Dominik
    Aug 16, 2015 at 12:29
  • 1
    $\begingroup$ I don't have a reference, because the lemma is pretty much trivial. If $x \le y$ wouldn't hold, you would have $y < x$ and therefore a real number $z$ exists such that $y < z < x$. But this implies $z \le y$, which is a contradiction. $\endgroup$
    – Dominik
    Aug 16, 2015 at 18:09

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