Absolute convergence of $\sin(n)/(n^2)$ Prove that $$\sum_{n=1}^{\infty}  \frac{\sin(n)}{{n}^{2}}$$
is either absolutely convergent, conditionally convergent or divergent.
Note that $$\sin(n) \in [-1,1] \text { for} 
\left| \sum_{n=1}^{\infty} \frac{\sin(n)}{{n}^{2}} \right|$$ 
is bound between $ 0 \text { and }1$.
So we have
$$0\leq \frac{\sin(n)}{{n}^{2}} \leq \frac{1}{{n}^{2}}$$
$$\lim_{n\to\infty} \frac{1}{{n}^{2}}=0 $$
and since
$$\frac{\sin(n)}{{n^{2}}}$$
is bounded between 0 and 0, it converges. 
I haven't proved the non-absolute of the series but I'd like to know if I'm in the right direction.
Edit:
Dam what is going wrong with LaTeX. I don't know how I got into this mess. 

You may prove the absolute convergence of the series by writing
$$
\left|\sum_{n=1}^{\infty}  \frac{\sin(n)}{{n}^{2}}\right|\leq\sum_{n=1}^{\infty}  \left|\frac{\sin(n)}{{n}^{2}}\right|\leq\sum_{n=1}^{\infty}  \frac{1}{{n}^{2}}
$$ thus your series is convergent.
