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

  • $\begingroup$ 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. $\lvert \sum_{k=1}^n \sin(k^2 x) \vert \leq M(x)$? $\endgroup$
    – AlanD
    Jul 25, 2021 at 3:21
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    $\begingroup$ 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? $\endgroup$ Jul 25, 2021 at 12:24

1 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).

  • $\begingroup$ 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? $\endgroup$ Jul 25, 2021 at 13:12
  • $\begingroup$ 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. $\endgroup$ Jul 25, 2021 at 13:16
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    $\begingroup$ 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. $\endgroup$
    – metamorphy
    Jul 25, 2021 at 13:23
  • $\begingroup$ Alright, thanks again for your help - I'm more than grateful! It has probably saved me days of trying to show something that isn't the case:) $\endgroup$ Jul 25, 2021 at 13:25

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