# Radius of convergence, prove that $\sum\limits_{n=0}^{\infty} a_n z^n$ converges absolutely for every $z \in \mathbb{C}$ with $|z| < R$

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.

Proof:

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 $$R$$?

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

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 $z_N$.
• Thank you. When you say $|z_n| \nearrow R$, I'm guessing you mean that $|z_n|$ tends to $R$ from below? Dec 19 '14 at 12:38