# convergence/divergence problem

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?

• How do you conclude that "it doesn't work" ? – Yves Daoust Oct 18 '17 at 9:05
• Are the limits on the sum correct now, after Jack's edit? – Kevin Oct 18 '17 at 9:07
• Limits are now correct yes. – Saeed Mohanna Oct 18 '17 at 9:09
• $\sin(n^2)$ 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. – Jack D'Aurizio Oct 18 '17 at 9:11

## 1 Answer

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 $n > N$.

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