If the radius of convergence is defined as $R$ such that the power series in $x$ (centered at $0$) converges for $|x|<R$ and diverges for $|x|>R$, I would like a proof that this $R$ exists. As far as I can tell, it boils down to the following statement:
If the power series $\sum a_n x^n$ converges at $x_0\in\mathbb C$, then it converges (absolutely) for any $x\in\mathbb C$ such that $|x|<|x_0|$.
Is that statement correct? If so, how can we prove it? One of the answers on this question mentions something about comparison to a geometric series, but I don't quite see what the answerer is talking about.
Best Answer
Yes the statement is correct. To see this: since $\sum a_n x_0^n$ converges so the sequence $(a_nx_0^n)$ is bounded say by $M$ (convergent to $0$) so
$$x_0\ne0\qquad |a_n x^n|=|a_n x_0^n|\left|\frac{x}{x_0}\right|^n\le M\left|\frac{x}{x_0}\right|^n$$ and for $0\le|x|<|x_0|$ the geometric series $\displaystyle \sum \left|\frac{x}{x_0}\right|^n$ is convergent.