Series absolute convergence

What could you say about absolute convergence of this series? $$\sum_1^\infty \frac{\sin(n) \sin(n^2)}{n}$$

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You want to go from $n=1$, not $0$. I think it does not converge absolutely. Note that $|\sin(n)\sin(n^2)| \ge \sin^2(n) \sin^2(n^2) = \frac{1}{4} - \frac{1}{4} \cos(2n) + \frac{1}{8} \cos(2n^2-2n) - \frac{1}{4} \cos(2n^2) + \frac{1}{4} \cos(2n^2+2n)$. Now $\sum_n \frac{1}{4n}$ diverges, $\sum_n \frac{1}{4n} \cos(2n)$ converges. If, as I suspect, $\sum_n \cos(2n^2 + 2jn)/n$ converges for $j=-1,0,1$, that shows that your sum does not converge absolutely. – Robert Israel Dec 19 '11 at 7:10
ah, got the hitch. Deleting my answer. – NikBels Dec 19 '11 at 7:26
@RobertIsrael: $\sum \cos(2n^2+2jn)/n$ converges by arguments similar to those in math.stackexchange.com/questions/2270/… – David Speyer Dec 19 '11 at 15:10
Because $\sum{sin(n) \over n}$ converges, multiplying each term by a number x such that 0<x<1 should yield a series which also converges. – Ben Sep 26 '12 at 0:36
@Ben Only if by "converges", you mean "converges absolutely". – Did Jan 4 '13 at 15:11

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

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This is amazing! – Pedro Tamaroff Feb 10 '13 at 20:50
@PeterTamaroff: It's a pretty simple idea, just a bit of computation. – robjohn Feb 10 '13 at 20:57
I am amazed by the simplicity, yes. – Pedro Tamaroff Feb 10 '13 at 21:29

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.

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