# Evaluating $\sum\frac{\sin(n)}{n^a}$

I have a function defined as: $$S(a)=\sum_{n=1}^\infty\frac{\sin(n)}{n^a}$$ My question is for what values of $$a$$ is this convergent, and how can I evaulate this?

For starters, I know that $$S(a)$$ is definately convergent for a $$a\ge2$$ since: $$\sum_{n=1}^\infty\frac{\sin(n)}{n^a}\le\sum_{n=1}^\infty\frac{1}{n^a}$$ So we can use $$\zeta(a)$$ as the upper limit for the series, which makes me think I can extend this convergence to $$a>1$$. However, I have no idea how I would evaluate this. Thanks!

• Exactly: the series converges absolutely for $\;a>1\;$ (I regard you mean $\;a\in\Bbb R\;$)...at least. – DonAntonio May 17 at 15:22
• This comes very close to math.stackexchange.com/q/3105348. – Paul Frost May 17 at 15:52

Hint: The series converges for all $$a>0$$ by Dirichlet's test as $$f(N)=\sum_{n=1}^N \sin(n)$$ is bounded and $$g(n)=\frac{1}{n^a}$$ is monotonically decreasing to zero.
Hint: for $$a>1$$ $$\sum_{n=1}^\infty\frac{\sin(n)}{n^a}<\sum_{n=1}^\infty\frac{|\sin(n)|}{n^a}$$ and $$\sum_{n=1}^\infty\frac{|\sin(n)|}{n^a}<\sum_{n=1}^\infty\frac{1}{n^a}$$