TITLE: Convergence of $\sum_{k=1}^\infty \frac{\sin(k(k-1))}{k}$
QUESTION [6 upvotes]: In connection with the strange behavious of a certain sum in Mathematica (https://mathematica.stackexchange.com/q/210849/16361) I suspected a possible divergence but I could not prove of disprove it.
Here's the question: is the sum 
$$s_1=\sum_{k=1}^\infty \frac{\sin\left(k(k-1)\right)}{k}$$
convergent or divergent?
Similarly with
$$s_2=\sum_{k=1}^\infty \frac{\sin(k^2)}{k}$$
Numerical evidence (partial sums) seem to indicate convergence.
EDIT 07.12.19
Actually, the story began one step earlier: I considered this unanswered question Convergence of $\sum_{n=1}^{\infty} \frac{\sin(n!)}{n}$ 
$$s_3 =\sum_{k=1}^\infty \frac{\sin(k!)}{k}$$
and wanted to simplify it replacing $k!$ with something simpler.

REPLY [3 votes]: Instead of proving convergence, I'll try calculating the limit.
For some large positive integer $n$ we have
$$
s_1 = \sum_{k=1}^{\infty} \frac{\sin(k(k-1))}{k}
= \sum_{k=1}^n \frac{\sin(k(k-1))}{k} + \sum_{k=n+1}^{\infty} \frac{\sin(k(k-1))}{k}
$$
The first term we can calculate exactly. For the second term we assume that $k(k-1)\mod2\pi$ is distributed uniformly in the interval $[0,2\pi)$ and so the sine functions can be replaced by independent random variables with zero mean and a variance of $\frac12$. We find the mean value
$$
\langle s_1\rangle = \sum_{k=1}^n \frac{\sin(k(k-1))}{k} + \sum_{k=n+1}^{\infty} \frac{\langle\sin(k(k-1))\rangle}{k}= \sum_{k=1}^n \frac{\sin(k(k-1))}{k}
$$
and the total variance
$$
\text{Var}(s_1) = \sum_{k=n+1}^{\infty} \frac{\frac12}{k^2}
= \frac12\psi'(n+1)=\frac{1}{2n}+O(n^{-2}).
$$
We thus find the physics-style estimate
$$
s_1 = \left[\sum_{k=1}^n \frac{\sin(k(k-1))}{k}\right] \pm \frac{1}{\sqrt{2n}}
$$
for $n\to\infty$.
Using $n=5\times10^7$ terms in the first sum, I find $s_1=0.3128±0.0001$.
In the same way, $s_2=0.1667±0.0001$.
More generally, we can try to plot the function
$$
f(z) = \sum_{k=1}^{\infty} \frac{\sin(k(k+z))}{k}
$$
where $s_1=f(-1)$ and $s_2=f(0)$: this function seems to have a lot of structure,