Series absolute convergence

Suppose that $|\sin(n^2)|\le a$ and $|\sin(n^2+2n+1)|\le a$, then $$ \begin{align} |\sin(2n+1)| &=|\sin(n^2+2n+1)\cos(n^2)-\cos(n^2+2n+1)\sin(n^2)|\\ &\le2a\tag{1}\\ |\sin(4n+2)| &\le|2\sin(2n+1)\cos(2n+1)|\\ &\le4a\tag{2} \end{align} $$ Therefore, $$ \begin{align} |\sin(2n+3)| &=|\sin(2)\cos(2n+1)+\sin(2n+1)\cos(2)|\\ &\ge|\sin(2)|\sqrt{1-4a^2}-2a|\cos(2)|\tag{3}\\ |\sin(4n+8)| &=|\sin(6)\cos(4n+2)+\sin(4n+2)\cos(6)|\\ &\ge|\sin(6)|\sqrt{1-16a^2}-4a|\cos(6)|\tag{4} \end{align} $$ Thus, $$ \begin{align} &|\sin(n^2+4n+4)|\\ &\qquad=|\sin(2n+3)\cos(n^2+2n+1)+\sin(n^2+2n+1)\cos(2n+3)|\\ &\qquad\ge|\sin(2n+3)|\sqrt{1-a^2}-a|\cos(2n+3)|\tag{5}\\ &|\sin(n^2+6n+9)|\\ &\qquad=|\sin(4n+8)\cos(n^2+2n+1)+\sin(n^2+2n+1)\cos(4n+8)|\\ &\qquad\ge|\sin(4n+8)|\sqrt{1-a^2}-a|\cos(4n+8)|\tag{6} \end{align} $$ Setting $a=0.047$ and using $(3)-(6)$, we get that if $$\left|\,\sin\left(n^2\right)\,\right|\le.047\quad\text{and}\quad\left|\,\sin\left((n+1)^2\right)\,\right|\le.047\tag{7} $$ then $$\left|\,\sin\left((n+2)^2\right)\,\right|\ge.841\quad\text{and}\quad\left|\,\sin\left((n+3)^2\right)\,\right|\ge.047\tag{8} $$ $(7)$ and $(8)$ imply

Lemma 1: only two of $\left|\,\sin\left(n^2\vphantom{1^2}\right)\,\right|$, $\left|\,\sin\left((n+1)^2\right)\,\right|$, $\left|\,\sin\left((n+2)^2\right)\,\right|$, and $\left|\,\sin\left((n+3)^2\right)\,\right|$ can be less than $0.047$.

Suppose that $|\sin(n)|\le b$, then $$ \begin{align} |\sin(n+1)| &=|\sin(1)\cos(n)+\sin(n)\cos(1)|\\ &\ge|\sin(1)|\sqrt{1-b^2}-b|\cos(1)|\tag{9}\\ |\sin(n+2)| &=|\sin(2)\cos(n)+\sin(n)\cos(2)|\\ &\ge|\sin(2)|\sqrt{1-b^2}-b|\cos(2)|\tag{10}\\ |\sin(n+3)| &=|\sin(3)\cos(n)+\sin(n)\cos(3)|\\ &\ge|\sin(3)|\sqrt{1-b^2}-b|\cos(3)|\tag{11}\\ \end{align} $$ Setting $b=0.070$ and using $(9)-(11)$, we get that if $$ |\sin(n)|\le.070\tag{12} $$ then $$ |\sin(n+1)|\ge.801\quad\text{and}\quad|\sin(n+2)|\ge.877\quad\text{and}\quad|\sin(n+3)|\ge.071\tag{13} $$ $(12)$ and $(13)$ imply

Lemma 2: only one of $|\sin(n)|$, $|\sin(n+1)|$, $|\sin(n+2)|$, and $|\sin(n+3)|$ can be less than $0.070$

The Lemmas 1 and 2 show that $|\sin(n)\sin(n^2)|\ge.00329$ at least once every four integers. This and comparison with the harmonic series imply that $$ \sum_{n=1}^\infty\left|\,\frac{\sin(n)\sin(n^2)}{n}\,\right|\tag{14} $$ diverges

Using the multi-dimensional Weyl-criterion you can prove that the tuple $((1/2\pi)n,(1/2\pi)n^2)$ is equidistributed in $\mathbb{T}^2$. Therefore for a positive Proportion of integers $n$ we will have $|\sin(n)\sin(n^2)| > \delta$, with some small fixed $\delta > 0$. Therefore the sum does not converge absolutely.