Problem
Define $$ f_1= \begin{cases} x & \text{if } x\leq \frac12\\ 1-x & \text{if }\frac12 \leq x \leq 1 \end{cases}$$ Then extend over $[0,\infty]$ by defining that for $x\geq 1$, $f_1(x+1) = f_1(x)$. Then define for $n\geq 2$, $f_n(x)=\frac12 f_{n-1}(2x)$. Let $S_m(x)=\sum_{n=1}^{m}f_n(x)$. Then $S_m$ is continuous on $[0,\infty)$ and the sequence $(S_m)_{m\in\mathbb{N}}$ converges uniformly to a continuous function $S$.
Show that $S$ is not differentiable at any point in $(0,\infty)$.
My Attempt
I am working through a textbook to practice real analysis in preparation for an upcoming analysis sequence. I have been stuck on this problem for hours and have no one to ask about it.
My first idea was to switch the limits:
$$ \lim_{h\rightarrow 0}\frac{S(x+h)-S(x)}{h}=\lim_{m\rightarrow\infty}\lim_{h\rightarrow 0}\frac{S_m(x+h)-S_m(x)}{h}$$.
Then I realized that $f_1$ isn't differentiable at multiples of $\frac12$, $f_2$ isn't differentiable at multiples of $\frac14$, and in general $f_n$ isn't differentiable at multiples of $\frac{1}{2^n}$.
Intuitively, as m approaches $\infty$ then $x$ gets closer and closer to some multiple of $\frac{1}{2^n}$ for some $n$. Hence, $f_m(x)$ approximately isn't differentiable at some m (I mean this in the vaguest of terms). If $f_m(x)$ isn't differentiable, then $S_m(x)$ isn't differentiable. Hence, $S(x)$ isn't differentiable. However, this is where I stuck. $\textbf{Does anyone know what to do next? Am I even on the right track?}$ Thank you!