What could you say about absolute convergence of this series? $$ \sum_1^\infty \frac{\sin(n) \sin(n^2)}{n} $$
Series absolute convergence
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calculus
sequences-and-series
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4You 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. – 2011-12-19
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0ah, got the hitch. Deleting my answer. – 2011-12-19
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0@RobertIsrael: $\sum \cos(2n^2+2jn)/n$ converges by arguments similar to those in http://math.stackexchange.com/questions/2270/convergence-of-sum-limits-n-1-infty-sinnk-n – 2011-12-19
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0Because $\sum{sin(n) \over n}$ [converges](http://math.stackexchange.com/questions/36732/is-the-sum-of-sinn-n-convergent-or-divergent), multiplying each term by a number x such that 0
– 2012-09-26 -
0@Ben Only if by "converges", you mean "converges absolutely". – 2013-01-04