Show that $ f(x)=\sum_{n=1}^{\infty} 2^{-n} f_n(x)$ defines a continuous function on $(0,\infty)$

Question

Let $f_n$ be a sequence of continuous functions on $(0,\infty)$ with $|f_n(x)|\le n$ for every $ x>0$ and $n\ge1$, and such that $\lim_{x\to\infty} f_n(x) =0$ for each $n$.Show that $ f(x)=\sum_{n=1}^{\infty} 2^{-n} f_n(x)$ defines a continuous function on $(0,\infty)$ that also satisfies $\lim_{x\to\infty} f(x) =0$. I am thinking of applying Weierstrass M test over here but I have hard time to figure out it. does Weierstrass M test relevant to my problem, if yes how?

Answer 1

Suppose $f_{n}$ is as given in the hypothesis of the problem. Suppose $\alpha \in \mathbb{R}$ such that $|\alpha| < 1$. The sequence of functions $g_{m}(x) = \sum_{n=1}^{m} \alpha^{n}f_{n}(x)$ is continuous. Since $\left|\alpha^{n}f_{n}(x) \right| \leq M_{n} = \alpha^{n}n$ and $\sum_{n=1}^{\infty} M_{n} = \frac{\alpha}{\left(1- \alpha \right)^{2}} < \infty$, by Weierstrass M-Test, $\{g_{m}\}$ is uniformly continuous. By the uniform limit theorem, $f(x) = \lim_{m\to \infty} g_{m}(x)$ must be continuous.