If $|x_1-x_0|I have a long question regarding the power series $\sum_{n=1}^\infty a_n(x-x_0)^n$.
Let $0 \le R \le \infty$ be the convergence radius of the power series $\sum_{n=1}^\infty a_n(x-x_0)^n$.
In the first question I have successfully proved that for $|q|<1$ the limit $\lim_{n \to \infty}[n \cdot q^n]=0$, while using the power series $\sum_{n=1}^\infty n \cdot x^n$ .
Now I have this question:
If $|x_1-x_0|<R$, then the series $\sum_{n=1}^\infty n \cdot a_n(x-x_0)^{n-1}$ converges absolutely at $x_1$.
Consider $q=\frac{|x_1-x_0|}{|x_2-x_0|}$ where $|x_1-x_0|<|x_2-x_0|<R$
I have tried to use the things I have proved above, but I don't have any clue on what to do.
I will appreciate some hints on how to prove this.
Thanks a lot!
 A: Since $\left|x_{2}-x_{0}\right|<R$, We know that $\displaystyle{{\displaystyle \sum_{n=N}^{\infty}a_{n}\left(x_{2}-x_{0}\right)^{n}}}$ converges absolutely. 
Let $q=\left|\frac{x_{1}-x_{0}}{x_{2}-x_{0}}\right|<1$. as you already showed, $\displaystyle{\lim_{n\to\infty}nq^{n}=0}$ ,that is, there exists $N\in\mathbb{N}$  such that $\forall n\ge N$ we have
$$\left|n\cdot\left(\frac{x_{1}-x_{0}}{x_{2}-x_{0}}\right)^{n}\right|<\left|x_{1}-x_{0}\right|\;\;\;\left(*\right)$$
Hence:
$$\sum_{n=N}^{\infty}\left|n\cdot a_{n}\left(x_{1}-x_{0}\right)^{n-1}\right|=\sum_{n=N}^{\infty}\left|n\cdot a_{n}\left(\frac{x_{1}-x_{0}}{x_{2}-x_{0}}\right)^{n}\cdot\frac{\left(x_{2}-x_{0}\right)^{n}}{x_{1}-x_{0}}\right|\underset{\left(*\right)}{\le}\sum_{n=N}^{\infty}\left|a_{n}\left(x_{2}-x_{0}\right)^{n}\right|<\infty$$
Namely, the power series converges absolutely at $x_1$
A: We have
$$
\sum\limits_{n = 1}^\infty  {n\left| {a_n } \right|\left| {x_1  - x_0 } \right|^{n-1} }  = \frac{1}{q}\sum\limits_{n = 1}^\infty  {nq^n \left| {a_n } \right|\left| {x_2  - x_0 } \right|^{n-1} } .
$$
You showed that $nq^n \to 0$. Thus there is an $N>0$ such that $nq^n <1$ for all $n\geq N$. Therefore,
\begin{align*}
\sum\limits_{n = N}^\infty  {n\left| {a_n } \right|\left| {x_1  - x_0 } \right|^{n-1}} & = \frac{1}{q}\sum\limits_{n = N}^\infty  {nq^n \left| {a_n } \right|\left| {x_2  - x_0 } \right|^{n-1} } \\ & \le \frac{1}{q|x_2-x_0|}\sum\limits_{n = N}^\infty  {\left| {a_n } \right|\left| {x_2  - x_0 } \right|^n }  <  + \infty .
\end{align*}
since the original power series is absolutely convergent for $|x-x_0|<R$. Thus, the radius of convergence is at least $R$. It is easy to see that it cannot be larger than that.
