# Proving that the radius of convergence of a power series is $r=1/(\lim \sup \sqrt[n]{|a_n|})$

Let $r$ the radius of convergence of the power series $\sum a_n(x-x_0)^n.$ Prove that if $r\in \mathbb{R^+}$, then $r=1/L$, where $L$ is the greatest subsequential limit of the bounded sequence $(\sqrt[n]{|a_n|})$. Hence, $r=1/(\lim \sup \sqrt[n]{|a_n|}$).

Any help on how to solve this? In class it was presented to us that $r=1/L$, where $L = \lim \sqrt[p]{|a_n|}$. This result supposes that the sequence $(\sqrt[p]{|a_n|})$ converges, and if so it will be in fact equal to $\lim \sup$. But how to proceed with more generality, without assuming the convergence of the series in order to prove that $r=1/(\lim \sup \sqrt[n]{|a_n|}$)?

We define $$r:= \frac{1}{\limsup_{n \rightarrow \infty} |a_n|^{1/n}}.$$ Let $0 \leq r_0< r$, then we can find $\epsilon >0$ with $r_0 +\epsilon < r$. By definition of $r$ we know that there exists $N \in \mathbb{N}$ with $$\sup_{k \geq N} |a_k|^{1/k} < (r_0+\epsilon)^{-1}.$$ Thus, for any $|x| < r_0$ we have for all $n \geq N$ $$|x|^n |a_n| \leq \Big( \frac{r_0}{r_0+\epsilon}\Big)^n.$$ Since $\delta = r_0/(r_0+\epsilon) <1$ series comparison test shows that the power series convergence uniformly in $|x| < r_0$. Thus, the convergence radius is at least $r$.
On the other hand, for any $r_0 > r$, we get that there exists a sequence $(n_k)_{k \in \mathbb{N}}$ of natural numbers such that $$|a_{n_k}|^{1/n_k} > (r_0 -\epsilon)^{-1}.$$ Thus $|a_{n_k}| |x|^{n_k}| \geq r_0/(r_0-\epsilon) > 1$. Therefore, the series is divergent. This implies that the convergence radius is precisely $r$.
Note that we cannot say anything about the convergence for $|x|=r$. For example $$-ln(1-x) = \sum_{k=1}^\infty \frac{x^k}{k}$$ has radius of convergence $1$. For $x=1$ this series is not convergent, but for any $x= \exp(2 \pi i \phi)$ with $\phi \in (0,1)$ it is convergent and equals $-\ln(1-x)$.