Prove that any power series converges uniformly on any compact set inside the circle of convergence.

Question

I am reading "Lectures on Complex Function Theory" by Takaaki Nomura.

There is the following problem in this book.

I read the answer of the problem.

But I cannot understand why it is sufficient to show that the power series converges uniformly on $\{z \in \mathbb{C} | |z| \leq r\}$ for any $r$ such that $0 < r < \rho$. I think we need to prove that any compact set inside the circle of convergence is a subset of $\{z \in \mathbb{C} | |z| \leq r\}$ for some $r$ such that $0 < r < \rho$.

And I cannot understand why we see the power series converges uniformly on $\{z \in \mathbb{C} | |z| \leq r\}$ from the last inequality.

Problem:

Prove that any power series $\sum_{n=0}^\infty a_n z^n$ converges uniformly on any compact set inside the circle of convergence.

Answer:

Let $\rho > 0$ be the radius of convergence.
It is sufficient to show that the power series converges uniformly on $\{z \in \mathbb{C} | |z| \leq r\}$ for any $r$ such that $0 < r < \rho$.
Let $r$ be a real number such that $0 < r < \rho$.
Let $\delta$ be a real number such that $r < r + \delta < \rho$.
By Cauchy - Hadamard theorem, $\frac{1}{\limsup_{n \to \infty} (|a_n|)^{\frac{1}{n}}} = \rho$.
So, there exists a natural number $N$ such that if $n > N$, then $(|a_n|)^{\frac{1}{n}} < \frac{1}{r+\delta}$.
If $|z| \leq r$ and $n > N$, then $|a_n z^n| \leq (\frac{|z|}{r+\delta})^n \leq (\frac{r}{r+\delta})^n$.
From this inequality, we see the power series converges uniformly on $\{z \in \mathbb{C} | |z| \leq r\}$ since $\frac{r}{r+\delta} \in (0, 1)$.

Answer 1

I think "inside the circle of convergence" means contained the open disk, centred at the centre of the power series, with radius equal to the radius of convergence. That is, with centre $a$ and radius $R$, we're looking at compact subsets of the open set $$\{z \in \Bbb{C} : |z - a| < R\}.$$ If this is the case, then it indeed suffices to consider compact subsets of $$\{z \in \Bbb{C} : |z - a| \le r\}$$ where $0 < r < R$. Why? The extreme value theorem. If $K \subseteq \{z \in \Bbb{C} : |z - a| < R\}$ is compact, then the function $z \mapsto |z - a|$ is continuous, and hence must achieve a maximum. This maximum $r$ must be strictly less than $R$ in order to remain in $K$ and hence $\{z \in \Bbb{C} : |z - a| < R\}$. Thus, $$K \subseteq \{z \in \Bbb{C} : |z - a| \le r\},$$ where $0 < r < R$.

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As for the second question, the proof has found a number $\lambda = \frac{r}{r + \delta} \in (0, 1)$, which does not depend on $z$ (!), such that $$|a_n z^n| \le \lambda^n.$$ So, if we consider the remainder series: $$\sum_{k=0}^\infty a_k z^k - \sum_{k=0}^n a_k z^k = \sum_{k=n+1}^\infty a_k z^k,$$ then $$\left|\sum_{k=n+1}^\infty a_k z^k\right| \le \sum_{k=n+1}^\infty |a_k z^k| \le \sum_{k=n+1}^\infty \lambda^k = \frac{\lambda^{n+1}}{1 - \lambda}.$$ Therefore, the remainders, when subtracting partial sums from the full sum, must approach $0$ at least as fast as the sequence $\frac{\lambda^{n+1}}{1 - \lambda}$, which is independent of $z$. Therefore, the series converges uniformly.