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Does the following series absolutely converge, conditionally converge or diverge?

$$\sum_{n=1}^{+\infty}\sin(n^2)\sin\left(\frac{1}{n^2}\right)$$

I don't even know where to begin, I tried the limit comparison test with $b_n= 1/n^2$, but it does not work.

What should I do?

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  • $\begingroup$ How do you conclude that "it doesn't work" ? $\endgroup$ – Yves Daoust Oct 18 '17 at 9:05
  • $\begingroup$ Are the limits on the sum correct now, after Jack's edit? $\endgroup$ – Kevin Oct 18 '17 at 9:07
  • $\begingroup$ Limits are now correct yes. $\endgroup$ – Saeed Mohanna Oct 18 '17 at 9:09
  • $\begingroup$ $\sin(n^2)$ is bounded between $-1$ and $1$, $\sin\frac{1}{n^2}$ behaves like $\frac{1}{n^2}$ for large values of $n$, hence the given series is absolutely convergent. Nothing tough here. $\endgroup$ – Jack D'Aurizio Oct 18 '17 at 9:11
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Comparison works !

$\frac{\sin(1/n^2)}{1/n^2} \to 1$ for $n \to \infty$. Hence there is $N$ such that

$\sin(1/n^2) \le 2/n^2$ for $n > N$. Therefore

$|\sin(n^2)\sin\left(\frac{1}{n^2}\right)| \le 2/n^2$ for $n > N$.

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  • $\begingroup$ Ahhhh i forgot about the direct comparison test, thanks Fred! So the answer would be absolutely convergent right? $\endgroup$ – Saeed Mohanna Oct 18 '17 at 9:11
  • $\begingroup$ Yes, you are right ! $\endgroup$ – Fred Oct 18 '17 at 9:16

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