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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!

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

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Hint: for $a>1$ $$\sum_{n=1}^\infty\frac{\sin(n)}{n^a}<\sum_{n=1}^\infty\frac{|\sin(n)|}{n^a}$$ and $$\sum_{n=1}^\infty\frac{|\sin(n)|}{n^a}<\sum_{n=1}^\infty\frac{1}{n^a}$$

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