[Math] Absolute convergence of $\sin(n)/(n^2)$

Question

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.

Answer 1

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.