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.

Answer 2

A real or complex series $\sum\limits_{n=0}^\infty a_n$ is said to converge absolutely if $\sum\limits_{n=0}^\infty \left|a_n\right| = l $ for some $l\in \mathbb{R}$.

Observe that $\sum\limits_{n=1}^{\infty} \left|\frac{\sin(n)}{{n}^{2}}\right|\le \sum\limits_{n=1}^{\infty} \frac{1}{{n}^{2}}$. But we know that $\sum\limits_{n=1}^{\infty} \frac{1}{{n}^{2}}$ is convergent. Therefore by the comparison test $\sum\limits_{n=1}^{\infty} \frac{\sin(n)}{{n}^{2}}$ is convergent absolutely. That is $\sum\limits_{n=1}^{\infty} \frac{\sin(n)}{{n}^{2}}$ is convergent.