A proof of this is given in my lecture notes as follows:

We define $R$ to be $\sup \{|z| \in \mathbb{R} : \sum |c_k z^k|$ converges $\}$ when the supremum exists.

Prove that $\sum |c_k z^k|$ converges for $|z| < R$ where $R$ is the radius of convergence.


Fix $z$ with $|z| < R$ and pick $S$ such that $|z| < S < R$. Then $\epsilon = R − S > 0$ and by Approximation Property for supremum we can find $\rho$ with $R−\epsilon = S < \rho < R$ such that $\sum |c_k \rho ^k|$ converges. But since $|z| < \rho$, this implies $\sum |c_k z^k|$ converges, by the simple Comparison Test.

I don't understand two things.

Firstly, why do they pick $S$ and then find $\rho$? Can't they just pick an $S$ that satisfy the extra conditions of $\rho$, that is, choose an $S$ that satisfies $R - \epsilon < S < R$ by the Approximation Property?

Secondly, why can they they say that there will definitely be a value between $R- \epsilon$ and $R$, calling it $\rho$, such that $\sum |c_k \rho ^k|$ converges. Why isn't it possible that no such value exists between $R - \epsilon$ and $R$?

  • $\begingroup$ No \displaystyle in titles, please. $\endgroup$
    – Did
    Dec 19 '14 at 13:11
  • $\begingroup$ $\rho = R - \epsilon/2$ is between $R-\epsilon$ and $R$… $\endgroup$
    – Dirk
    Dec 19 '14 at 13:21
  • $\begingroup$ @Did okay, thanks for telling me. May I ask why? $\endgroup$
    – Elise
    Dec 19 '14 at 13:59
  • $\begingroup$ Unnecessary formatting, vertical space taken to others, as explained everywhere. $\endgroup$
    – Did
    Dec 19 '14 at 15:10
  • $\begingroup$ @Did Okay, thank you. I should have realised. $\endgroup$
    – Elise
    Dec 19 '14 at 15:16

I hope this helps a little bit:

  1. $R - \varepsilon < S < R$ cannot be, since $\varepsilon = R - S$. However, I'm not quite sure why they dont just pick a $\rho$ such that $R - |z| < \rho < R$. I guess it's just to show the thought process: Since $|z| < R$ there exists a $S$ such that $|z| < S < R$. Afterwards, we can choose $\rho$ between $S$ and $R$ with the desired properties.

  2. I believe it should be $R - \varepsilon <\rho\leq R$. The existence of such a $\rho$ can be proven as follows: Since $R = \sup \{ |z| \in \Bbb R ~ : ~ \sum |c_kz^k| \text{ converges } \}$, there exists a sequence $(z_n)$ of complex numbers such that $\sum |c_k z_n^k|$ converges for all $n \in \Bbb N$ and $|z_n| \nearrow R$. By definition of the limit, it follows that there exists some $N\in \Bbb N$ such that $$R - |z_N| < \varepsilon$$ and hence, $$ R - \varepsilon < |z_N| \le R.$$ Now we define $\rho := |z_N|$ and get $$\sum |c_k \rho^k| = \sum |c_k| |\rho|^k = \sum |c_k| |z_N|^k = \sum |c_k z_N^k|$$ which converges by our choice of $z_N$.

  • $\begingroup$ Thank you. When you say $|z_n| \nearrow R$, I'm guessing you mean that $|z_n|$ tends to $R$ from below? $\endgroup$
    – Elise
    Dec 19 '14 at 12:38
  • $\begingroup$ @Elise That's right $\endgroup$
    – j4GGy
    Dec 19 '14 at 12:39

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