TITLE: Series absolute convergence
QUESTION [13 upvotes]: What could you say about absolute convergence of this series?
$$
\sum_1^\infty \frac{\sin(n) \sin(n^2)}{n}
$$

REPLY [3 votes]: 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