Determine convergence of series I want to detemine the convergence of the next series:
$$\sum_{n=1}^\infty\frac{\sin(na)}{n^2} $$
I've solved the limit:
$$\lim_{n->\infty}\frac{\sin(na)}{n^2}=\frac{[ -1,1]}{\infty}=0$$
The series has the necesary condiction of convergence (limit=0), but I dont go further from here
 A: Show that it’s absolutely convergent, i.e., that $$\sum_{n=1}^\infty\left|\frac{\sin na}{n^2}\right|$$ converges. You should have no trouble doing this by comparing it with a simple series that you know converges.
Note, by the way, that it makes no sense at all to write $$\lim_{n\to\infty}\frac{\sin na}{n^2}=\frac{[-1,1]}\infty\;:$$ $\infty$ isn’t something by which you can divide, and $[-1,1]$ isn’t even a number. What you mean is this:

For all $n$, $$\frac{-1}{n^2}\le\frac{\sin na}{n^2}\le\frac1{n^2}\;,$$ so $$0=\lim_{n\to\infty}\frac{-1}{n^2}\le\lim_{n\to\infty}\frac{\sin na}{n^2}\le\lim_{n\to\infty}\frac1{n^2}=0\;,$$ and therefore by the squeeze theorem $$\lim_{n\to\infty}\frac{\sin na}{n^2}=0\;.$$

A: $-1 \leq \sin(na) \leq 1$ and hence
$$\left|  \frac{sin(na)}{n^2} \right| \leq \frac{1}{n^2}$$
Since $\sum \frac{1}{n^2}$ is convergent....
A: Use the comparison test. Since $\left|\sin(na)\right| \le 1$, we have:
$$
\left| \frac{\sin(na)}{n^2} \right| \le \left| \frac{1}{n^2} \right|
$$
We know that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges absolutely, so your series does too.
A: We get absolute convergence by the comparison test since
$$
\sum_{n=1}^{\infty} |\frac{\sin na}{n^2}| \leq \sum_{n=1}^{\infty} \frac{1}{n^2},
$$
and $\sum_{n=1}^{\infty} \frac{1}{n^2}$  converges. Since the series converges absolutely, it converges. This holds for every real number $a$.
A: We can use direct comparison of $$\sum_{n=1}^\infty\left|\frac{\sin(na)}{n^2}\right|$$ with the $2$-harmonic series $$\sum_{n=1}^\infty\frac{1}{n^2}.$$ The latter converges, so the former is absolutely convergent, and so convergent.
