Let $(x)=x-n$, where $n$ is the unique integer such that $n\leq x<n+1$. I need to prove that $f(x)=\sum_{n=1}^\infty \frac{(nx)}{n^2}$ is discontinuous at all rational numbers.
Here is my approach: Let $x=\frac pq$ be a rational where $p$ and $q$ are relatively prime. Let us rearrange the sum as follows: Let $n=kq+r$, where $0\leq r<q$. Then $f(x)=\sum_{r=0}^{q-1}\sum_{k=0}^\infty \frac {((kq+r)x)}{(kq+r)^2}$. Now it is easy to see that $\sum_{k=0}^\infty \frac {((kq+r)x)}{(kq+r)^2}$ is continuous at $x=\frac pq$ for $1\leq r<q$ since $((kq+r)\frac pq)$ is continuous at $x=\frac pq$ because $(kq+r)\frac pq=kp + \frac {rp}q$ where $\frac {rp}q$ is clearly not an integer as $p$ and $q$ are coprime. Since the series converges uniformly, the limit is continuous at each point where all the terms are continuous. In particular, $\sum_{k=0}^\infty \frac {((kq+r)x)}{(kq+r)^2}$ is continuous at $x=\frac pq$ for $1\leq r<q$.
Now all I need to show is that $\sum_{k=0}^\infty \frac {((kq+r)x)}{(kq+r)^2}$ is discontinuous at $x=\frac pq$ for $r=0$. I have no idea how to prove it.