How do we prove that a power series centered at $a\in\textbf{R}$ is continuous on the interval $(a-R,a+R)$?

Question

Let $f:(a-R,a+R)\to\textbf{R}$ be the function \begin{align*} f(x) = \sum_{n=0}^{\infty}c_{n}(x-a)^{n} \end{align*} For any $0 < r < R$, the series $\displaystyle\sum_{n=0}^{\infty}c_{n}(x-a)^{n}$ converges uniformly to $f$ on the compact interval $E = [a-r,a+r]$.

In particular, $f$ is continuous on $(a-R,a+R)$.

My solution

Let us define $f_{n}(x) = c_{n}(x-a)^{n}$ on $E$. Then each $f_{n}$ is a continuous and bounded real-valued function. Hence we have that \begin{align*} \|f_{n}\| = \sup_{x\in E}|f_{n}(x)| = |c_{n}|r^{n} \Rightarrow \sum_{n=0}^{\infty}\|f_{n}\| = \sum_{n=0}^{\infty}|c_{n}|r^{n} = \sum_{n=0}^{\infty}|c_{n}r^{n}| \end{align*}

Since $0 < r < R$, the last numerical series converges.

Consequently, due to the Weierstrass $M$-test, the proposed power series converges uniformly on $E$, whence $f$ is continuous.

My question is: how do I prove from the obtained results that $f$ is continuous on the open interval $(a-R,a+R)$?

Answer 1

Take $x\in(a-R,a+R)$. Then $|x-a|<R$. Take some $r\in\bigl(|x-a|,R\bigr)$. Then $f$ is continuous in $[a-r,a+r]$ and, in particular, in $(a-r,a+r)$, to which $x$ belongs. So, $f$ is continuous at $x$.

Answer 2

hint

Take $x_0$ in $(a-R,a+R)$.

Assume that $a<x_0<a+R$.

Put $$y=x_0+\frac{a+R-x_0}{2}$$ then

$$x_0<y<a+R$$

$f$ is continuous at $[-y,y]$, so it is continuous at $x_0$.