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

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

• Regarding your edit, if you applied Dirichlet's test, the proof makes use of the boundedness of $\sum_{k=1}^n b_k,$ and then uses a comparison test. If you were successful here, did your bound depend on $x$, but not $n$? i.e. $\lvert \sum_{k=1}^n \sin(k^2 x) \vert \leq M(x)$? Jul 25, 2021 at 3:21
• That's actually a very good point! I'll check my work in the Dirichlet test. However, if I follow this convergence proof for the convergence of $\sum_{n=1}^{\infty} \frac{\sin(n^2)}{n}$: math.stackexchange.com/questions/809346/… , I think I should be able to claim that x is simply a scaling factor and convergence is preserved for every case except for $x = \pi/2$, correct? Jul 25, 2021 at 12:24

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).
• Hi, thanks so much for that! Let me try to rephrase it just to make sure I actually understand (sorry - a lot of the concepts in number theory are still somewhat foreign to me). What this states is that if $x = 2\pi p / q$ where $p$ and $p$ are coprime, the sine ($\sin(n^2x)$) is periodic with q - so $\sin(n^2x) = \sin((n+q)^2x)$. Now if the sum from the sines between $1$ and $q$ is not zero, that means something is continually added to the total sum for every period of q --> divergence. This does not tell me anything about other points where $x \neq 2\pi p / q$ though, correct? Jul 25, 2021 at 13:12
• I'm aware the the last question is outside of the scope of the question but interesting to analyse the initial fractal (say I can show the at least for the points where $x \neq 2 \pi p / q$ there exists a derivative that probably allows me to make some statement about the upper bound of my fractal dimension. Jul 25, 2021 at 13:16
• The density breaks any hope for uniform convergence anywhere (so this approach of taking the derivative fails). The convergence at $x$ being some other rational multiple of $2\pi$ may be analysed using the "generalised" version of the Gauss sum (as found in the Wiki page); for instance, it holds for prime $q\equiv 1\pmod 4$. But the scope is up to you. Jul 25, 2021 at 13:23