Is $\sum_{n=1}^{\infty} \frac{\sin(n^2x)}{n}$ uniformly convergent on $(0,\pi)$?

Question

I am trying to prove that the function $\sum_{n=1}^{\infty} \frac{\sin(n^2x)^2}{n^3}$ is not a fractal by showing that it has a well defined derivative (as fractals do not). In order to do that, I have to find out whether the function $\sum_{n=1}^{\infty} \frac{\sin(n^2x)}{n}$ is uniformly convergent on the interval $(0,\pi)$. If it is, the original function is not a fractal!

It is clear that using the Weierstrass M-test it can be shown that: $\sum_{n=1}^{\infty} \frac{\sin(n^2x)}{n^\alpha}$ where $\alpha > 1$ is uniformly convergent since $\sum_{n=1}^{\infty} \frac{1}{n^\alpha}$ converges and $|\frac{\sin(n^2x)}{n^\alpha}| \leq \frac{1}{n^\alpha}$.

Now the case where $\alpha = 1$ the function $\sum_{n=1}^{\infty} \frac{\sin(nx)}{n}$ (no $n^2$ in the sine) is the fourier trasnform of a sawtooth wave - so it converges uniformly everywhere except for when $x$ is a multiple of $\pi$. I'm not sure if the function I'm investigating (with $n^2$ in the sine) would share a similar property.

I have done quite a bit of research and it seems nobody has analysed this specific function yet and I'm a bit unsure as to how I can continue here. I believe that somehow the following substitution might help:

$$\sum_{n=1}^{\infty} \frac{\sin(n^2x)}{n} = \sum_{n=1}^{\infty} \frac{1}{2in} (e^{i n^2 x} - e^{-i n^2 x})$$

But I can't get to any results from here either. It would be amazing if you could give me some pointers as I'm making no progress (I'm a non-math PhD student who is stuck figuring this out) and am wasting ungodly amounts of time on this without a solution in sight.

Thanks so much for your help in advance!

EDIT: It can be proven that $\sum_{n=1}^{\infty} \frac{\sin(n^2x)}{n}$ is pointwise convergent using Dirichlet's test fairly easily. <--- This is incorrect - there was a mistake in my derivation

Answer 1

In fact, this series diverges on a dense subset of $(0,\pi)$. (Consider $x=\pi/2$ first.)

Let $x=2\pi p/q$ be a rational multiple of $2\pi$ (with $p\perp q$). Then the map $n\mapsto s_n:=\sin(n^2 x)$ is $q$-periodic (that is, $s_{n+q}=s_n$) and then, denoting $\bar{s}:=(1/q)\sum_{n=1}^q s_n$, we know that $\sum_{n=1}^\infty s_n/n$ converges if and only if $\bar{s}=0$ (the "only if" part follows from $s_n=\bar{s}+(s_n-\bar{s})$ and the "if" part which, in turn, is shown using Dirichlet's test).

But, if $q$ is a prime with $q\equiv 3\pmod{4}$ (enough for the density), then $$\sum_{n=1}^q s_n=\Im\sum_{n=1}^q e^{2i\pi n^2 p/q}=\left(\frac{p}{q}\right)\sqrt{q}\neq 0,$$ where $\left(\frac{p}{q}\right)$ is the Legendre symbol (see Quadratic Gauss sum on Wikipedia).