A confusion about the radius of convergence in Ahlfors' "Complex Analysis"?

Question

On the third edition of Ahlfors' Complex Analysis, page 39 Theorem 2 it states: The derived series $\sum_{1}^{\infty}na_n z^{n-1}$ has the same radius of convergence, because $\sqrt[n]n \rightarrow 1$.

I'm confused about it.

The radius of convergence of $\sum_{0}^{\infty}a_nz^n$ is $\lim_{n\rightarrow \infty} \sup \sqrt[n]{|a_n|}$

We know that \begin{equation*} \sum_{1}^{\infty}na_nz^{n-1}=\sum_{0}^{\infty}(n+1)a_{n+1}z^n=\sum_{0}^{\infty} b_nz^n \end{equation*} So, the radius of convergence of $\sum_{1}^{\infty}na_nz^{n-1}$ should be $\lim_{n\rightarrow \infty} \sup \sqrt[n]{|b_n|}=\lim_{n\rightarrow \infty} \sup \sqrt[n]{|(n+1)a_{n+1}|}$

If we want to prove these two power series have the same radius of convergence, we should prove:

(i) $\sqrt[n]{n+1} \rightarrow 1$. But I'm confused that Ahlfors says he wants $\sqrt[n]n \rightarrow 1$.

(ii) $\lim_{n\rightarrow \infty} \sup \sqrt[n]{|a_n|} = \lim_{n\rightarrow \infty} \sup \sqrt[n]{|a_{n+1}|}$. But I don't know how to write a rigorous proof for this.

Answer 1

Suppose that the power series $\sum_{n=0}^\infty a_n x^n$ has radius of convergence $R$. Rewrite the derived series as: $$\sum_{n=1}^\infty na_n x^{n-1}\stackrel{k=n-1}{=}\sum_{k=0}^\infty \underbrace{(k+1)a_{k+1}}_{b_k}x^k$$ and let $R'$ be its radius of convergence. We want to show that $R'=R$. Thanks to Cauchy-Hadamard theorem: $$\begin{align*}\frac{1}{R'}&=\limsup_{n\to\infty}|b_n|^\frac{1}{n}\\&=\limsup_{n\to\infty}|(n+1)a_{n+1}|^\frac{1}{n}\\&=\limsup_{n\to\infty}\underbrace{(n+1)^\frac{1}{n}}_{\alpha_n}\underbrace{|a_{n+1}|^\frac{1}{n}}_{\beta_n}=\color{red}*\end{align*}$$ At this point: $$\begin{align*}\lim_{n\to\infty}\alpha_n&=\lim_{n\to\infty}(n+1)^\frac{1}{n}\\&=\lim_{n\to\infty}e^\frac{\log(n+1)}{n}\\&=e^{\underbrace{\lim_{n\to\infty}\frac{\log(n+1)}{n}}_{=0}}\\&=1\end{align*}$$ In general, given two real and positive sequences $(a_n)_n$ and $(b_n)_n$ , if $\lim_{n\to\infty}a_n$ exists then: $$\limsup_{n\to\infty}a_nb_n=\lim_{n\to\infty}a_n\limsup_{n\to\infty}b_n$$ This allows us to conclude that: $$\color{red}*=\limsup_{n\to\infty}\beta_n$$ Let's compute it: $$\begin{align*}\limsup_{n\to\infty}\beta_n&=\limsup_{n\to\infty}|a_{n+1}|^{\frac{1}{n+1}\frac{n+1}{n}}\\&=\limsup_{n\to\infty}e^{\frac{n+1}{n}\log|a_{n+1}|^\frac{1}{n+1}}=\color{blue}\blacktriangle\end{align*}$$ Since the exponential is a continuous, nondecreasing and positive real function we know that: $$\begin{align*}\color{blue}\blacktriangle&=e^{\limsup_{n\to\infty}\frac{n+1}{n}\log|a_{n+1}|^\frac{1}{n+1}}\\&=e^{\limsup_{n\to\infty}\log|a_{n+1}|^\frac{1}{n+1}}\\&=\limsup_{n\to\infty}e^{\log|a_{n+1}|^\frac{1}{n+1}}\\&=\limsup_{n\to\infty}|a_{n+1}|^\frac{1}{n+1}\\&=\frac{1}{R}\end{align*}$$ Thus we have finally shown that $R'=R$.