Series calculus

See here .

Since $\displaystyle |(-1)^n\frac{\sin(n)}{n^2}| = |\frac{\sin(n)}{n^2}| \leq \frac{1}{n^2}$ , and since $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ converges to $\displaystyle\frac{\pi^2}{6}$ , then by the comparison test $\displaystyle \sum_{n=1}^{\infty}(-1)^n\frac{\sin(n)}{n^2}$ must absolutely converge, too.

No and Not at all are same, Maybe is never the correct answer. But I want to know how to solve it.