Discontinuities of $f(x)=\sum \limits_{n=1}^{\infty}\frac{\{nx\}}{n^2}$ Let $f(x)=\sum \limits_{n=1}^{\infty}\dfrac{\{nx\}}{n^2}$ where $\{\}$ is fractional part. Find all discontinuities of function $f(x)$.

I think that $f(x)$ is discontinuous at every rational point. 

Can anyone show how to prove strictly that $f$ for example is discontinuous at $0$ or $1/2$?
Expanding the comment:

Let, $x = \dfrac{p}{q} \in \mathbb{Q}^{+}$ (the case of negative rationals can be dealt with similarly), and for a $n \in \mathbb{N}$, we have:

Either $nx \notin \mathbb{N}$, in which case for any positive $\epsilon$-nbd of the point $x$, where, $\displaystyle 0 &lt; \epsilon &lt; \frac{1}{n}$,

$$\left\{\frac{np}{q}\pm n\epsilon\right\} = \left\{\frac{np}{q}\right\} \pm n\epsilon$$

Else, if $nx \in \mathbb{N}$, for points in the right $\epsilon$-neighborhood of $x$,

$$\left\{\frac{np}{q} + n\epsilon\right\} = n\epsilon$$

and, for points in the left $\epsilon$-neighborhood of $x$,

$$\left\{\frac{np}{q}- n\epsilon\right\} = 1- n\epsilon$$

If we let $\epsilon \to 0$, the left and right limits when $nx \notin \mathbb{N}$ are the same, but they are not the same when $nx \in \mathbb{N}$.

Thus, the $N^{th}$ partial sum function: $\displaystyle f_N(x) := \sum\limits_{n=1}^{N} \frac{\{nx\}}{n^2}$, is discontinuous at all $x \in \mathbb{Q}$ for sufficiently large $N$ (we need only find a $N$ large enough so that $nx \in \mathbb{N}$ for some $n &lt; N$).

This jump discontinuity at any $x \in \mathbb{Q}$ is a positive number, call it: 

$\displaystyle J_N(x) = \left|\lim\limits_{\epsilon \to 0^+} f_N(x) - \lim\limits_{\epsilon \to 0^-} f_N(x)\right| &gt; 0$ is bounded, 

and since the tail $\displaystyle f(x) - f_N(x) = \sum\limits_{n=N+1}^{\infty} \frac{\{nx\}}{n^2} \to 0$ as $N  \to \infty$, we may let $N$ large enough to ensure that $J_N(x) &gt; |f(x) - f_N(x)|$, so that the contribution from the tail  can't make up for the jump at a discontinuity. Thus, $f(x)$ is always discontinuous at rational points.

Edit:

To prove uniform-continuity of $f_N$ at $x \in \mathbb{R \setminus Q}$,

If, $x \in \left(\frac{m}{N},\frac{m+1}{N}\right)$, then for any positive $\epsilon$ such that, $(x - \epsilon, x+\epsilon) \subset \left(\frac{m}{N},\frac{m+1}{N}\right)$ and $y \in (x - \epsilon, x+\epsilon)$,

$$|f_N(y) - f_N(x)| = \left|\sum\limits_{n=1}^{N} \frac{\{nx\} - \{ny\}}{n^2}\right| \le \epsilon \sum\limits_{n=1}^{N}\frac{1}{n^2} &lt; 2\epsilon$$
Start with $x = a$ and conclude that $f(x) = f(a)$.  What if we are not sure of the value of $x$.  Is it true that $x \approx a$ implies $f(x) \approx f(a)$.  This is the basis of our concept of "continuity" - often we are able to guess the value of a function based on "nearby" or "similar values".

Continuous functions behave nicely under limits: $\displaystyle \lim_{x \to a} f(x) = f(a)$ 

Let's use this definition to try to check your function is continuous for irrational $x \not \in \mathbb{Q}$ such as $x = \sqrt{\color{#0055FF}{2}}, \sqrt{3}$.  If I just write out the equations.

$$ \lim_{x \to \sqrt{\color{#0055FF}{2}}}\; \sum \limits_{n=1}^{\infty}\frac{\{nx\}}{n^2} \stackrel{?}{=} \sum \limits_{n=1}^{\infty}\frac{\{n \sqrt{\color{#0055FF}{2}}\}}{n^2} $$ 

I have no idea about the number on the right side.  Is it even finite?  It's smaller than another number I am familiar with.  Even if the function is discontinuous or not, at east point it has a finite value.

$$ \sum \limits_{n=1}^{\infty}\frac{ \{ n \sqrt{2}\}}{n^2} \leq \sum \limits_{n=1}^{\infty}\frac{ 1}{n^2}  = \frac{\pi^2}{6} &lt; \infty $$

What about nearby points?  Can we show that for $x \approx \sqrt{2}$ this function attains close values?

$$ \sum \limits_{n=1}^{\infty}\frac{ \{ n (\sqrt{2} \pm \color{red}{\epsilon})\}}{n^2} \stackrel{?}{\approx}  \sum \limits_{n=1}^{\infty}\frac{ \{ n \sqrt{2} \}}{n^2} $$

Let $\epsilon \ll 1$ be a really small number.  How about $\epsilon = \frac{1}{1000}$ or $\frac{1}{M}$.  We can bound the error in one of two possible ways.


$ \{ n \sqrt{2}  \} - n \epsilon &lt; \{ n \sqrt{2} \pm \epsilon \} &lt; \{ n \sqrt{2}  \} + n \epsilon$
the error here is smaller than $\frac{1}{n^2} \times n \epsilon = \frac{\epsilon}{n} \leq \epsilon$


Jumping does not occur before the $M = \frac{1}{\epsilon}$ term. For $n &gt; M$:  $ \displaystyle 0 \leq  \frac{ \{ n \sqrt{2} \pm \epsilon \}}{n^2} \leq \frac{1}{M^2} = \epsilon^2$.

It seems we have established uniform convergence in the terms of the series and therefore our limit is correct for $x \notin \mathbb{Q}$.





Here's the plot I found.   

It certainly looks discontinuous at $x = \frac{1}{2}$ as we expected.  It looks like a kind of "swoosh" shape repeated many times,  but a similar jump discontinuity should appear at all $x \in \mathbb{Q}$.

It's a totally reasonable-looking function. It's defined at all points in $[0,1]$, $\int_0^1 f(x) \, dx$ makes perfect sense and yet this function has horrible discontinuities everywhere!