Prove if $\displaystyle\sum_{n=0}^{\infty}a_n(x-x_o)^n$ is conditionally convergent at $x_1$, then $R=|x_1-x_0|$ ( $R$ is radius of convergence)
First, According to Abel's Theorem. I know that:
(1) if $\displaystyle\sum_{n=0}^{\infty}a_n(x-x_o)^n$ is convergent at $x_1$, then $R\ge|x_1-x_o|$
(2) if $\displaystyle\sum_{n=0}^{\infty}a_n(x-x_o)^n$ is divergent at $x_1$, then $R\le|x_1-x_o|$
Actually my text book says that the question can be proved using (1) and (2). And the hint given by the textbook is method of contradiction. This maybe a stupid question, cause this should be a part of Abel's Theorem. But I want to figure out how does (1) and (2) get my question.