Upper bound for modulus of the coefficients of a power series

Question

On p.56 of the fourth edition of Serge Lang´s complex analysis book, he states that if a power series $\sum a_n z^n$ has radius of convergence $R>0$, then there exist a constant $C>0$ such that if $A>1/R$, then $|a_n|\leq C A^n $ for all $n$, which is equivalent to say that for all $\varepsilon>0$ we have $$|a_n|\leq \frac{C}{(R-\varepsilon)^n}$$ for all $n$. He says that this follows from the proof that $1/R=\limsup |a_n|^{1/n}$ but i dont know why, ¿Could someone explain to me how this follows from the above result?

Answer 1

The radius of convergence $R$ of a power series $\sum_na_nz^n$ is given by the reciprocal of $\rho:=\limsup_n\sqrt[n]{|a_n|}$, that is $R=\frac{1}{\rho}$. If $0<\rho<\infty$, and $A>\frac{1}{R}=\rho$ then $A=\rho+\delta$ for some $\delta>0$. Then, by definition of $\limsup$, there exists $N\in\mathbb{N}$ such that $$\sqrt[n]{|a_n|}<\rho+\delta\quad\text{whenever}\quad n\geq N$$ This implies that for $n\geq N$, $|a_n|\leq \big(\rho+\delta)^n=A^n$. Let $C=\max\Big(1,\max\limits_{1\leq n<N}\frac{|a_n|}{A^n}\Big)$. Then $$|a_n|\leq CA^n,\qquad n\in\mathbb{Z}_+$$

Now $R>\frac{1}{A}$ and so, setting $\varepsilon=R-\frac1A>0$, we have that $\frac{1}{A}=R-\varepsilon$ and so, $A=\frac{1}{R-\varepsilon}$. Putting things together we have that $$|a_n|\leq \frac{C}{(R-\varepsilon)^n}, \qquad n\in\mathbb{Z}_+$$

The case $\rho=0$ ($R=\infty$) is left to the OP.