Prove that the series $f(x) = \sum_{n=1}^{\infty}2^n \sin(\frac{1}{3^nx})$ defines a continuous function on $(0, +\infty)$

Question

As the title says, I want to prove the continuity of the limit function of such series. I know that if the serie converges absolutely on an interval then it is continuous there. Also, the series is absolutely convergent on $[a, +\infty), a > 0$, as per this post: study the uniform convergence of this sereis of functions:

I'm having trouble understanding why this proves that the limit function is continuous on the whole interval $(0, +\infty)$. Is this just because any $x \in (0, +\infty)$ is in $[a, +\infty)$ for some $a \in \mathbb{R}$?

Answer 1

Roughly: yes.

More precisely: $f$ is continuous at any point $x_0>0$ because $x_0\in(a,+\infty)$ for some $a>0$ (e.g. $a=\frac{x_0}2$). It is important to take $a<x_0$ and not just $a\le x_0,$ to make sure $f$ is continuous at $x_0$ and not just right-continuous.