Absolute convergence of power series, true/false? Is the following statement true? If so, how can I prove it?

If the power series $\sum_{n=0}^{\infty} a_n x^n$ converges for all $x
 \in (x_0, x_1)$ then it converges absolutely for all $x \in
 (-\max\{|x_0|, |x_1|\}, \max\{|x_0|, |x_1|\})$.

 A: The trick is first showing absolute convergence. Once you show that, then for any $y_0 \in (-\max\lbrace |x_0|, |x_1| \rbrace, \max\lbrace |x_0|, |x_1| \rbrace)$, we can find a $y_1 \in (x_0, x_1)$, with $|y_1| > |y_0|$. Then, because of absolute convergence at $y_1$, we can use the comparison test:
$$\sum |a_n| |y_0|^n \le \sum |a_n| |y_1|^n < \infty,$$
so $\sum a_n y_0^n$ converges absolutely.
Now to show absolute convergence. Take an arbitrary $y_0 \in (x_0, x_1)$ and choose $y_1$ in the same interval such that $|y_1| > |y_0|$.  Then, since $\sum a_n y_1^n$ converges, by the divergence test, we must have $a_n y_1^n \rightarrow 0$. It therefore follows that there exists an $N$ such that
$$n \ge N \implies |a_n| |y^n_1| < 1 \implies |a_n| |y^n_0| < \left|\frac{y_0}{y_1}\right|^n.$$
Note that $|y_0/y_1| < 1$, so we have a convergent series, so by the comparison test, $\sum a_n y_0^n$ converges absolutely.
A: All power series are based on this
$$0 <a <b\implies 0 < \frac {a}{b}<1$$
and
$|q|<1\implies  \sum q^n $ converges.
The rest is artistic.
