Why if a power series $\displaystyle\sum^{\infty}_{n=0}a_n(x-x_0)^n$ converges in $x=x_1$, then it it absolutely convergent when $|x-x_0|<|x_1-x_0|$?
and if the power series diverges in $x=x_1$, it is divergent when $|x-x_0|>|x_1-x_0|$?
Could we make a comparison with a geometric series? If yes, how could I do that?
Suppose that $x_1\ne x_0$ and the series $\sum_{n=0}^\infty a_n(x_1-x_0)^n$ converges. Then, from the ratio test
$$\begin{align} \lim_{n\to \infty}\left|\frac{a_{n+1}(x_1-x_0)^{n+1}}{a_{n}(x_1-x_0)^{n}}\right|&=|x_1-x_0|\,\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\\\&=L\\\\&\le 1 \end{align}$$
Therefore, the limit of the ratio of successive coefficients is
$$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{L}{|x_1-x_0|}$$
Then, using the ratio test on the convergence of the series $\sum_{n=0}^\infty a_n(x_2-x_0)^n$, where $0<|x_2-x_0|<|x_1-x_0|$, we have
$$\begin{align} \lim_{n\to \infty}\left|\frac{a_{n+1}(x_2-x_0)^{n+1}}{a_{n}(x_2-x_0)^{n}}\right|&=|x_2-x_0|\,\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\\\&=\frac{|x_2-x_0|}{|x_1-x_0|}L\\\\&\le \frac{|x_2-x_0|}{|x_1-x_0|}\\\\&<1 \end{align}$$
and the series $\sum_{n=0}^\infty a_n(x_2-x_0)^n$ also converges as was to be shown!