continuous series

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!

I would say your argument is unnecessarily complicated instead of being sloppy. As you have pointed out, by Weierstrass test and the theorem that the uniform limit of a sequence of continuous maps is continuous we can prove that $f$ is continuous:
By assumption we have $$\Bigg| \frac{f_{n}(x)}{2^{n}} \Bigg| \leq \frac{1}{2^{n}}$$ for all $x > 0$ and all $n \geq 1$; moreover, we have $$\frac{1}{2^{n}} = \frac{1}{\exp (n\log 2)} < \frac{1}{n^{2}}$$ for large $n$, so by comparison test the series $\sum_{n \geq 1}\frac{1}{2^{n}}$ converges; hence by Weierstrass test the series $f$ converges uniformly and then is continuous.
• Seems to me that it is easier to prove $\sum 1/2^n$ converges since partial sums are easy to get. – marty cohen Oct 27 '15 at 2:28