Convergence of a sine series

Question

Using Mathematica, we claim that the following series is convergent: $$\sum_{n=1}^{\infty}\frac{\sin(n^2 t)}{n}$$ Any idea how we prove this?

Answer 1

Let $t=\frac{\pi}{2}$.

Then since the sequence of squares mod $4$ is $$\{1^2,2^2,3^2,4^2,5^2,6^2,7^2,8^2,\ldots\}\equiv\{1,0,1,0,1,0,1,0,\ldots\}$$ your series will be the same as $$\begin{align} &\frac{\sin(\pi/2)}1+\frac{\sin(0)}2+\frac{\sin(\pi/2)}3+\frac{\sin(0)}4+\frac{\sin(\pi/2)}5+\frac{\sin(0)}6+\frac{\sin(\pi/2)}7+\frac{\sin(0)}8+\cdots\\ &=\frac11+\frac13+\frac15+\frac17+\cdots \end{align}$$ which is divergent.

So at least for $t=\pi/2$, this is actually not a convergent series.

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In general, for $t=2\pi/q$, if the squares mod $q$ are not balanced between "highs" and "lows", it seems likely that there will be divergence. For example, with $q=7$, the sequence of squares is $\{0,1,4,2,2,4,1,\text{repeat}\}$. So you have $$\begin{align} &\sum_{n=0}^{\infty}\left(\frac{\sin(2\pi/7)}{7n+1}+\frac{\sin(8\pi/7)}{7n+2}+\frac{\sin(4\pi/7)}{7n+3}+\frac{\sin(4\pi/7)}{7n+4}+\frac{\sin(8\pi/7)}{7n+5}+\frac{\sin(2\pi/7)}{7n+6}\right)\\ &=\sum_{n=0}^{\infty}\left(\sin(4\pi/7)\left(\overbrace{\frac{2\cos(4\pi/7)}{7n+2}+\frac{1}{7n+3}}^{\text{positive}}+\overbrace{\frac{1}{7n+4}+\frac{2\cos(4\pi/7)}{7n+5}}^{\text{positive}}\right)+\frac{\sin(2\pi/7)}{7n+1}+\frac{\sin(2\pi/7)}{7n+6}\right)\\ &>\sum_{n=0}^{\infty}\left(\frac{\sin(2\pi/7)}{7n+1}+\frac{\sin(2\pi/7)}{7n+6}\right) \end{align}$$ which is divergent.

So it seems like there are lots of $t$ for which this does not converge. Having said that, for $t$ that are not rational multiples of $\pi$, my guess (it's only a guess) would be that the sum converges, since the sign of the terms would exhibit pseudorandom behavior.