# Discontinuous for rationals

Show that $f\left(x\right):=\sum_{n=1}^{\infty}\frac{\left\{nx\right\}}{n^2}$, where $\left\{nx\right\}$ is the fractional part of $nx$, is discontinuous for all rationals.

I guess it would be nice to somehow show that for any arbitrarily small irrational $r$, $f\left(\frac{a}{b}\right)<f\left(\frac{a}{b}+r\right)$, but I am not sure how to do that precisely (though it makes sense intuitively since $\left\{n\frac{a}{b}\right\}=0$ for all $n=mb$ for some $m\in\mathbb{N}$ whereas $\left\{n\left(\frac{a}{b}+r\right)\right\}\ne0\,\forall\,n\in\mathbb{N}$.)

Out of curiosity I tried beginning by computing the value of $f\left(x\right)$ for $x\in\mathbb{Q}$. My analysis showed that if $x=\frac{a}{b}$, where (without loss of generality) $\frac{a}{b}\in\left[0,\,1\right]$ and (again, without loss of generality) $\gcd\left(a,\,b\right)=1$ then: $$f\left(\frac{a}{b}\right)=\frac{a}{b^3}\sum_{l=1}^{b-1}l\frac{d}{dx}\left(\ln\left(\Gamma\left(\frac{l}{b}\right)\right)\right)$$ But I don't think that's relevant.

Note: this question comes from Rudin's Principles of Mathematical Analysis Exercise 7.10.

For $l,\, m \in \mathbb{N}$ consider: \begin{aligned} f_{N} \left( {\frac {l}{m}}+{\frac {\epsilon}{m}} \right) &=\sum _{n=1}^{ N } \frac{\left\{n \left( {\frac {l}{m}}+{\frac {\epsilon}{m} } \right) \right\}}{n^2}, \quad |n\frac{\epsilon}{m}|<1 \end{aligned} \tag{1} Firstly, when $$n\frac{l}{m}\notin \mathbb{N} \tag{2}$$ then: $$\left\{n \left( {\frac {l}{m}}\pm{\frac {\epsilon}{m} } \right) \right\}=\left\{n {\frac {l}{m}} \right\}\pm n\frac {\epsilon}{m}\tag{3}$$ but consider what happens when: $$n\frac{l}{m}=k,\quad k\in \mathbb{N} \tag{4}$$ For a positive perturbation: $$\left\{n \left( {\frac {l}{m}}+{\frac {\epsilon}{m} } \right) \right\}=\left\{ k+n\frac{\epsilon}{m}\right\}=n\frac{\epsilon}{m}$$ while for a negative perturbation $$\left\{n \left( {\frac {l}{m}}-{\frac {\epsilon}{m} } \right) \right\}=\left\{ k-n\frac{\epsilon}{m}\right\}=1-n\frac{\epsilon}{m} \tag{5b}$$ Similarly, when: $$n\frac{l}{m}=-k,\quad k\in \mathbb{N} \tag{6}$$ then for a positive perturbation: $$\left\{n \left( {\frac {l}{m}}+{\frac {\epsilon}{m} } \right) \right\}=\left\{ -k+n\frac{\epsilon}{m}\right\}=-1+n\frac{\epsilon}{m} \tag{7a}$$ while for a negative perturbation $$\left\{n \left( {\frac {l}{m}}-{\frac {\epsilon}{m} } \right) \right\}=\left\{ -k-n\frac{\epsilon}{m}\right\}=-n\frac{\epsilon}{m} \tag{7b}$$
When we take $\epsilon \rightarrow 0$, the limit is the same from both sides in $(3)$ but not in $(5)$ and $(7)$. Consequently, we have shown that $\forall N\in\mathbb{N}$, $f_{N}(x)$ is discontinuous at $x\in \mathbb{Q}$. Now consider: $$f(x)=f_{N}(x)+\sum_{N+1}^{\infty}\frac{\left\{n x\right\}}{n^2} \tag{8}$$ As we choose ever larger $N$, the value of the jump at the discontinuity in $f_N(x)$ will tend to some none zero value, call it $d_{N}$, while the tail, call it $T_{N}$ (sum on the right in $(8)$), becomes ever smaller. It follows that $\exists N$ such that $d_N>T_N$ in which case the discontinuity can never be saved by the tail and it follows that $f(x)$ is discontinuous at $x\in\mathbb{Q}$.