Prove that if the power series $\displaystyle \sum_{n=0}^\infty a_nx^n$ converges for $x = x_0$, then it converges absolutely for all $x$ such that $|x| <|x_0|$.
I am supposed to use the following corollary to the Root Test to prove the above prompt:
COROLLARY: Let $\{a_n\}$ be a sequence of real numbers. Then,
(a) if $\limsup_{n\to\infty} \sqrt[n]{|a_n|} = 0$, then the series $\displaystyle \sum_{n=0}^\infty a_nx^n$ converges absolutely for all real $x$.
(b) if $\limsup_{n\to\infty} \sqrt[n]{|a_n|} = L > 0$, then the series $\displaystyle \sum_{n=0}^\infty a_nx^n$ converges absolutely for $|x| <{1\over L}$ and diverges for $|x| > {1 \over L}$.
(c) if $\limsup_{n\to\infty} \sqrt[n]{|a_n|} = \infty$, then the series $\displaystyle \sum_{n=0}^\infty a_nx^n$ converges only for $x = 0$.
Here is my attempt at proving the prompt:
PROOF:
For (a), if we suppose that $\limsup_{n\to\infty}\sqrt[n]{|a_n|} = 0$, then by the corollary, we know that the power series converges absolutely for all real $x$, and by extension converges for all $x$ such that $|x| < |x_0|$.
For (b), if we suppose that $\limsup_{n\to\infty}\sqrt[n]{|a_n|} = L > 0$, note that from the Root Test, we have that $$\limsup_{n\to\infty}\sqrt[n]{|a_nx_0^n|} = |x_0|\limsup_{n\to\infty}\sqrt[n]{|a_n|} = |x_0|L < 1 \implies |x_0| < {1\over L},$$ since the power series is to converge when $x = x_0$. Hence for all real $x$ such that $|x| < |x_0|$, by the corollary, $|x| < |x_0| < {1\over L}$ gives us that $\displaystyle \sum_{n=0}^\infty a_nx^n$ will converge absolutely for all such $x$.
For (c), if $\limsup_{n\to\infty}\sqrt[n]{|a_n|} = \infty$, we know by the corollary that the power series will converge only for $x=0$ and the result is therefore trivial since there exists no $x \in \mathbb R$ such that $|x| < |x_0| = 0$. Thus we are done.
Here's my question: I'm a little confused as to if I really proved anything at all in (c). It seems like I encountered an impossibility rather than proved the prompt. Did I make a mistake here (or anywhere else for that matter)?
Best Answer
Your above proof is correct.
FRED