The book Understanding Analysis by Stephen Abbott asserts that
$ g(x)=\sum_{n=0}^{\infty}\frac{1}{2^n}h(2^nx), $
where $h(x)=\left|x\right|$, $h:\left[-1,1\right]\to\mathbb{R}$, and $h(x+2)=h(x)$, is continuous on all of $\mathbb{R}$ but fails to be differentiable at any point.
However, if I am not mistaken, can the $2^n$'s be cancelled out in $g$? I tried plotting this and could not obtain a nowhere-differentiable function.