Where is $f(x) := \sum_{n=1}^\infty \frac{\langle nx\rangle}{n^2+n}$ discontinuous? 
Let $\langle x\rangle$ denote the fractional part of $x\in
 \mathbb{R}$, i.e. $\langle x\rangle := \inf \{ x-k: k\in \mathbb{Z},
 k\leq x \}$. 
  
  
*
  
*Define $$f(x) := \sum_{n=1}^\infty \frac{\langle 
     nx\rangle}{n^2+n}.$$ What is the domain of $f$?
  
*Is $f$ Riemann integrable on any bounded interval? If so, evaluate
  $\int_0^1 f(x)dx$.
  
*Find all points where $f$ is discontinuous.
  

I'm fine with #1 and #2...I think the answer to #3 should be $\mathbb{Q}$, but this answer suggests that the function is continuous at the integers. I don't see any reason why it would be. What do you think?
 A: The "fractional part" function
$$
\{\,\} : \mathbb{R} \to [0,1),\quad x \mapsto \{ x\}:=x-\lfloor x\rfloor
$$
is obviously discontinuous at every integer point. 
If fact, since $\{\,\}$ is $1$-periodic, it is enough to show that it is discontinuous at $1$. For $\varepsilon>0$ small enough we have
\begin{eqnarray}
\{1+\varepsilon\}&=&1+\varepsilon-\lfloor 1+\varepsilon\rfloor=1+\varepsilon-1=\varepsilon\\
\{1-\varepsilon\}&=&1-\varepsilon-\lfloor 1-\varepsilon\rfloor=1-\varepsilon-0=1-\varepsilon.
\end{eqnarray}
Hence
$$
\lim_{x\to 1-}\{x\}=\lim_{\varepsilon\to0+}(1-\varepsilon)=1\ne0=\lim_{\varepsilon\to0+}\varepsilon=\lim_{x\to1+}\{x\}.
$$
If $n \in \mathbb{N}$, then the function 
$$
u_n:\mathbb{R} \to [0,1),\quad x\mapsto \{nx\}
$$
is discontinuous at every point of the form $\frac{k}{n}$, where $k \in \mathbb{Z}$.
Since
$$
f=\sum_{n=1}^\infty \frac{u_n}{n^2+n},
$$
it follows that $f$ is discontinuous at every point of the form $\frac{k}{n}$, where $n\in \mathbb{N}$, and $k\in \mathbb{Z}$, i.e. $f$ is discontinuous at every point of $\mathbb{Q}$.
A: Every irrational point is a point of continuity. Let $\alpha\in\Bbb R$ be irrational. Every function
$$f_n:\Bbb R\to \Bbb R, x\mapsto\frac{\langle nx\rangle}{n^2+n}$$ is continuous at $\alpha$. The series $$\sum f_n$$ is clearly normally convergent on $\Bbb R$, whence continuity at $\alpha$.

Every rational point is a point of discontinuity. Let $p/q$ be rational (with $q$ positive.) Fix $N>>q$ such that
$$\sum_{n> N}\frac{1}{n^2+n}<\frac{1}{2(q^2+q)}$$
The function
$$\sum_{1\leq n\leq N} f_n$$
experiences a discontinuity at $p/q$, specifically its value drops (at least) by $\frac{1}{q^2+q}$ when transitioning from $(p/q)^-$ to $(p/q)^+$. By definition of $N$, this drop is too big for the remainder
$$\sum_{n> N}f_n$$
to compensate for, and so $\sum f_n$ is discontinuous at $p/q$.
