If f is differentiable everywhere on $(0,1)$ we can say $$f(1)-f(0) =\int^1_0 f'(x)dx \ \ (*)$$ If we knew that $f$ is a monotone function, (*) would immediately imply that $f$ is AC. However we know that $f$ is of bounded variation, thus can be written as the difference of two monotonically increasing functions on $[0,1]$ : $$f(x)=[f(x)+{TV}(f_{[0,x]})] - TV(f_{[0,x]})$$ Where $TV(f_{[0,x]})$ is the total variation function, which is differentiable everywhere on $[0,1]$, since it is monotone on $[0,1]$ and by Lebesgue's theorem is differentiable.
The functions $f_n$ are obviously continuous on $]\frac{1}{n+1},\frac1n[$ (because here $f_n(x)$ is $\sin(\pi/x)$, which is continuous everywhere on $]0,\infty[$), and they are continuous on $[0,\frac{1}{n+1}[$ and $]\frac1n,\infty[$, because $f_n(x)=0)$ there$. Since
$$\lim_{x\to(\frac1{n})^+}f_n(x)=\lim_{x\to(\frac1{n})^-}f(x)=0$$
and likewise at $\frac1{n+1}$, $f_n$ is continuous everywhere.
For a given $x>0$, you can prove that $f_n(x)\to0$, simply because all the values of $f_n(x)$ are zero, except possibly finitely many. And $f_n(0)=0$ for all $n$. Hence $f_n$ converges "pointwise" to the zero function.
So, we have a sequence of continuous funtctions, that converges pointwise to the zero function, which is also continuous.
However, you are right, the maximum of $f_n(x)$ is always $1$, because for $x_n=\dfrac1{n+1/2}$, which is in $[1/(n+1),1/n]$,
$$\sin^2(\pi/x_n)=\sin^2(n\pi+\pi/2)=1$$
However, $f_n(x)$ converges uniformly if convergence happens at each point "at once", that is, for all $\epsilon>0$, there is some $n_0$ such that for all $x$ and all $n>n_0$, $|f_n(x)|<\epsilon$. It has to happen for instance for $\epsilon=1/2$. But this can't be true, since, with $x_n$ as above, $|f_n(x_n)|=1>\epsilon$.
Hence the sequence $f_n$ does not converge uniformly.
Now for $\sum_n f_n(x)$. The successive intervals $[1/(n+1),1/n]$ are disjoint except at $1/n$ and $1/(n+1)$, where $f_n$ is zero anyway, and their union is $]0,1]$, hence the sum is simply the function $f(x)=\sin(\pi/x)$ for $x\in]0,1]$ and $f(x)=0$ elsewhere. Your plot shows a part of it.
We have a theorem (see here, that states that if a series of continuous functions converges uniformly to a function $f$, then $f$ has to be continuous.
Here we have a sum of continuous functions, which converge to a function that is certainly not continuous at $0$: the limit $\lim_{n\to0^+}f(x)$ does not exists, and $f$ oscillates in every interval $[0,\epsilon[$. Hence the convergence can't be uniform.
Best Answer
This is false. The Cantor function $f(x)$ is a counterexample: it is uniformly continuous and of bounded variation, yet it is not absolutely continuous.
Uniformly continuous: $f$ is continuous on the compact set $[0,1]$
Bounded variation: $f$ is increasing and bounded
Not absolutely continuous: $f'=0$ almost everywhere, so $f(x)\not=\int_0^x f'(x)\ dx$.