Let $(f_n)$ be a sequence of continuous functions on $(0, \infty)$ satisfying $|f_n(x)| \leq 1$ for all $x > 0$ and all $n \geq 1$. Show that the function $f(x) = \sum_{n=1}^{\infty} \frac{f_n(x)}{2^n}$ defines a continuous function on $(0, \infty)$. If, in addition, the $f_n$ satisfy $\lim_{x \to \infty} f_n(x) = 0 $, show that $\lim_{ x \to \infty} f(x) = 0$, as well.
I have an idea of what to do: My idea is to couple the Weierstrass M-Test with the fact that the uniform limit of a sequence of continuous functions is continuous, for the first part; this should be sufficient for out result. I would think that second would easily follow from part 1.
Attempt:
Consider the sequence $(g_N) \subset C((0, \infty))$, where $g_N(x) = \sum_{n=1}^{N} 2^{-n} f_n(x)$ for each $N$. Clearly, the sequence $(g_N)$ converges pointwise to $f$. Since $|f_n(x)| \leq 1$ for all $x > 0$ and all $n \geq 1$, it follows that $||g_N||_{\infty} \leq ||\sum_{n=1}^{N} 2^{-n}||_{\infty}$. Thus, we see that $||f||_{\infty} = ||\lim_{n \to \infty} g_n||_{\infty} \leq \sum_{n=1}^{\infty} ||2^{-n}||_{\infty} = \sum_{n=1}^{\infty} 2^{-n}< \infty, $ since $\sum_{n=1}^{\infty} 2^{-n}$ is geometric. Hence, since $||f||_{\infty} < \infty$, by the Weierstrass M-Test, it must be that $(g_N)$ converges uniformly to $f$ on $(0, \infty)$. In particular, since each $g_N$ is continuous, it must be that $f$ is continuous as the uniform limit of continuous functions.
Do this argument hold water? I think my intuition is right, but the argument seems very sloppy to me. Any help is appreciated!
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