If $|_1−_0|>$, then $\lim_{n\to \infty}[a_n\cdot |x_1-x_0|^n]\neq 0$ Let $0\leq R \leq \infty$ be the convergence radius of the power series $\sum_{n=1}^\infty a_n(x-x_0)^n$.

Prove that if $|x_1-x_0| > R$, then $$\lim_{n\to \infty}[a_n\cdot |x_1-x_0|^n]\neq 0$$ hence we conclude that $\sum_{n=1}^\infty n\cdot a_n(x-x_0)^{n-1}$ diverges at $x_1$.

Consider $q=\frac{|x_2-x_0|}{|x_1-x_0|}$ where $|x_1-x_0|>|x_2-x_0|>R$, $lim_{n\to \infty}[n\cdot q^n]=0$.
I claimed that since $|x_1-x_0| > R$ then $\sum_{n=1}^\infty a_n(x_1-x_0)^n$ diverges.
Then I wanted to claim that the limit of summation is not 0, and keep going from there.
But I didn't know how to show that the limit of summation is not zero when the series doesn't converge.
Is there an easier method?
 A: I go with the following argument:
Assumption that $R$ is finite is required. Let us first assume that $R\neq 0$.  Now, by definition $$\lim\sup_{n\to \infty} |a_n|^{\frac{1}{n}} = \frac{1}{R}.$$ Hence $|x_1-x_0| \lim\sup_{n\to \infty} |a_n|^{\frac{1}{n}} = \frac{|x_1-x_0|}{R}>1$. In other words there exists a subsequence $\{|x_1-x_0||a_{n_k}|^{\frac{1}{n_k}}\}$ that converges to a value say $p>1$. So, $|x_1-x_0||a_{n_k}||^{\frac{1}{n_k}} >1$ for infinitely may $k$.  Hence $|x_1-x_0|^{n_k} |a_{n_k}| > 1$ for infinitely many $k$. This in turn implies that, $|a_n||x_1-x_0|^n$ cannot go to $0$.
If $R=0$, then definition of $R$ implies directly that there are infinitely many $k$ for which $|x_0-x_1||a_k|^{\frac{1}{k}}>1$. Hence the conclusion.
The other part follows easily.
A: I will prove the first part as I hope you can continue from there.
Let $x_{2}\in\mathbb{R}$ such that $\left|x_{1}-x_{0}\right|>\left|x_{2}-x_{0}\right|>R$ and set $q=\left|\frac{x_{2}-x_{0}}{x_{1}-x_{0}}\right|<1$. 
Now assuming by contradiction that $\displaystyle{\lim_{n\to\infty}\left[a_{n}\cdot\left|x_{1}-x_{0}\right|^{n}\right]=0}$, then there exists $N\in\mathbb{N}$ such that $\forall n\ge N$ we have:
$$\left|a_{n}\cdot\left|x_{1}-x_{0}\right|^{n}\right|<1\Longrightarrow\left|a_{n}\right|<\frac{1}{\left|x_{1}-x_{0}\right|^{n}},\;\;\;\left(*\right)$$
But that implies:
$$\sum_{n=N}^{\infty}\left|a_{n}\left(x_{2}-x_{0}\right)^{n}\right|\underset{\left(*\right)}{\le}\sum_{n=N}^{\infty}q^{n}\underset{\left|q\right|<1}{<}\infty$$
Which is a contradiction since $\left|x_{2}-x_{0}\right|>R$.
