Series that converge or diverge

Question

Let $\sum a_n$ with $a_n=\frac{\sin(n\theta)}{n^\alpha}$, $\theta\neq k\pi$, $k$ integer.

Want to prove that the series converges absolutely when alpha is strictly greater than $1$. Any hints please? what if $\alpha$ is between $0$ and $1$.

Answer 1

Start by noticing that for any $n\in\mathbb{N}$ and $\theta\in\mathbb{R}$, we know that $|\sin(n\theta)| \le 1$.

That way you can see that $$ \sum \left| \frac{\sin(n\theta)}{n^{\alpha}} \right| \le \sum \frac{1}{n^{\alpha}} $$

For $\alpha > 1$, $\sum \frac{1}{n^{\alpha}}$ definitely converges (that takes a little more to explain, so I'll leave it to you).

Since $\sum \frac{1}{n^{\alpha}}$ converges, then by the comparison test $\sum \left| \frac{\sin(n\theta)}{n^{\alpha}} \right|$ converges.

Thus $\sum \frac{\sin(n\theta)}{n^{\alpha}}$ converges absolutely, showing that it also converges.