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Is $\displaystyle\sum_{n=1}^{\infty}\frac{\sin\left(n^{2}x\right)}{n}$ convergent for all real $x$ or not?
If not, is it divergent everywhere except $x=n\pi,\ n\in\mathbb{Z}$?
Caution: It is not the same case as $\displaystyle\sum_{n=1}^{\infty}\frac{\sin\left(nx\right)}{\sqrt{n}}$.

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    $\begingroup$ Partial answer that I'm not sure where to post: If $x=\frac{\pi}{2}$ the numerator is $0$ at even numbers and $1$ at odd numbers, so it diverges. $\endgroup$ – Carl Schildkraut Dec 13 '17 at 14:56
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    $\begingroup$ If $x=\frac{2\pi}p$, where $p$ is an odd prime, the series converges iff $p\equiv1\pmod4$. (Hint: Quadratic Gauss sum) $\endgroup$ – Professor Vector Dec 13 '17 at 15:13
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As noted in a comment, it diverges for some $x$, for example $x=\pi/2$.

It does converge for almost every $x$, since it it the Fourier series of a periodic $L^2$ function.

That's by Carleson's Theorem on almost-everywhere convergence of Fourier series. That's a huge theorem, like the hardest theorem in Fourier analysis, but it does answer the question.

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    $\begingroup$ The almost-everywhere pointwise convergence of such series can be proved through summation by parts and Weyl's inequality, without invoking Carleson's theorem. If $x\approx\frac{p}{q}$, the partial sums of $\sin(n^2 x)$ are bounded by $O_q(\sqrt{n}\log n)$. $\endgroup$ – Jack D'Aurizio Dec 13 '17 at 19:01
  • $\begingroup$ @JackD'Aurizio Of course I knew there must be a better way - wasn't aware of any useful estimate on those partial sums. $\endgroup$ – David C. Ullrich Dec 13 '17 at 19:08

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