I am trying to solve the following problem:
We are given $A=\{a_1,a_2,a_3,...\}$, a countable set of real numbers that is bounded. Let $$ f(x)=\sum_{n=1}^\infty{\frac{|x-a_n|}{10^n}}.$$
Show that $f$ is convex on $\mathbb{R}$, differentiable on $\mathbb{R}\setminus A$, and nondifferentiable on $A$.
The book says that without using material involving sequences and series of functions (which it has not covered yet), we are able to solve this problem through definitions.
I have managed to show convexity. However, I do not know how to show differentiability. I tried writing out the definition for the right derivative: $$ f^\prime_+(x)=\lim_{h\to0+}\frac{\sum_{n=1}^\infty\frac{|x+h-a_n|}{10^n}-\sum_{n=1}^\infty\frac{|x-a_n|}{10^n}}{h} =\lim_{h\to0+}\sum_{n=1}^\infty\frac{|x+h-a_n|-|x-a_n|}{10^nh} $$
and since I had no idea how to proceed from there, I just used the triangle inequality in the following way: $$ f^\prime_+(x)\le\lim_{h\to0+}\sum_{n=1}^\infty\frac{|h|+|x-a_n|-|x-a_n|}{10^nh} =\lim_{h\to0+}\sum_{n=1}^\infty\frac{|h|}{10^nh} =\lim_{h\to0+}\sum_{n=1}^\infty\frac{1}{10^n}=\frac{1}{9} $$
I imagine doing the same thing to $f^\prime_-(x)$ will give a lower bound of $-\frac{1}{9}$. However, this still does not prove differentiability, since I need $f^\prime_+(x)=f^\prime_-(x)$. I also cannot figure out where the nondifferentiability at points of $A$ comes in.
Any guidance would be much appreciated.
