$\left(\frac{a_n}{n^k}\right)_n$ is bounded implies $\sum_{n=0}^\infty a_nz^n$ has a radius of convergence $\ge 1$

Question

Let $$\left(\frac{a_n}{n^k}\right)_n\subset\mathbb{C}\;\;\;\;\;(k\in\mathbb{N})$$ be a boundet sequence. I want to show that the power series $$\sum_{n=0}^\infty a_nz^n\;\;\;\;\;(a_n,z\in\mathbb{C})$$ has a radius of convergence $R\ge 1$

Proof:

• Since $\left(\frac{a_n}{n^k}\right)_n$ is bounded, it holds $$\left(\frac{a_n}{n^k}\right)_n\subset B_C(0)$$ for some $C>0$. Thus, $$|a_n|\le Cn^k\;\;\;\;\;\forall n$$
• So, we have $$\sqrt[n]{|a_n|}\le \sqrt[n]{Cn^k}=\sqrt[n]{C}\cdot n^\frac{k}{n}\to 1\cdot 1\;\;\;\text{for}\;\;\; n\to\infty$$ because $C$ is constant and with $$n^\frac{k}{n}=e^{\ln n^\frac{k}{n}}=e^{\frac{k}{n}\ln n}$$ and the fact $$\frac{k}{n}\ln n\to 0\;\;\;\text{for}\;\;\; n\to\infty$$ as well as the continuity of the exponential function
• From the root test it follows $$\sqrt[n]{|a_nz^n|}=\sqrt[n]{|a_n||z|^n}=\sqrt[n]{|a_n|}\cdot |z|\to |z|\;\;\;\text{for}\;\;\; n\to \infty$$
• Thus, the radius of convergence should be $R<1$

What am I doing wrong?

Answer 1

You proved that $\limsup \sqrt[n]{|a_n|} \le 1$. The radius of convergence is $R = \dfrac{1}{\limsup \sqrt[n]{|a_n|}}$.