Radius of convergence.

Question

let $(a_n)_{n\in\mathbb{N}}$ such that $a_n\neq 0$ for all $n\in\mathbb{N}$ and such that $P_n:=\sum_{k=0}^n a_kX^k$ splits over $\mathbb{R}$ and has no multiple root. I have to show that the radius of convergence of $\sum_{n=0}^{+\infty} a_nx^n$ is $+\infty$. There was an indication : first show that $$\sum_{\lambda\text{ root of }P_n}\frac{1}{\lambda^2}$$ is independent of $n$ for $n\geqslant 2$. I found out how to prove this but I don't know how to relate this to the initial problem. Any help would be appreciated.

Answer 1

By Cauchy-Hadamard theorem, the radius of convergence of the series $\{a_nx^n\}$ equals $\left(\lim \sqrt[n]{|a_n|}\right)^{-1}$. By Vieta’s formulae, $\left|\tfrac{a_0}{a_n}\right|=\left|\prod _{\lambda\text{ root of }P_n} \lambda\right|$ for each $n$. Thus $$|a_n|=|a_0|\cdot \left| \prod _{\lambda\text{ root of }P_n}\frac 1{\lambda} \right|\le |a_0|\left(\frac 1n \sum_{\lambda\text{ root of }P_n}\frac{1}{\lambda^2}\right)^{n/2},$$ which tends to zero when $n$ tends to infinity.