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I want to prove or disprove $\displaystyle\sum_{n\ge1} \frac{{\sin n}}{n}$ is absolutely convergent.

I can prove $\displaystyle \sum_{n\ge1} \frac{{\sin n}}{n}$ is convergent by dirichlet's test (Can I prove that $\displaystyle \sum_{n\ge1} \frac{{\sin n}}{n}$ converges without dirichlet's test), but I can't prove or disprove that $\displaystyle\sum_{n\ge1} \frac{\vert{\sin n}\vert}{n} $ converges.

Any help would be appreciated.

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    This was asked tons of times before. Roughly speaking, the idea is that in each triplet $$\{\sin(3n+1),\sin(3n+2),\sin(3n+3)\}$$ one of the sines is at least $\frac12$ or less than $-\frac12$, hence, for every $n$, $$\sum_{k=1}^3\frac{|\sin (3n+k)|}{3n+k}\geqslant\frac1{6(n+1)}$$ from which the divergence is direct. – 2017-02-07
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    Can I know max of $ \vert \sin{3n+1}\vert,\vert \sin{3n+2}\vert,\vert \sin{3n+3}\vert$ is greater than a half. – 2017-02-07
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    Sure, but you first: did you draw a trigonometric circle and try to understand why three successive angles cannot all be in the arcs where $|\sin|<\frac12$? What are these arcs, already? – 2017-02-07

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Hint:

$$\frac{| \sin n|}{n} \geqslant \frac{\sin^2 n}{n} = \frac{1}{2n} - \frac{\cos 2n}{2n}$$

$\sum \frac{\cos 2n}{2n}$ converges, $\sum \frac{1}{2n} $ diverges

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Hint: There are infinitely many integers which are more than $1$ away from the nearest multiple of $\pi$ (because in between every two consecutive multiples of $\pi$ there are three integers, and the middle one has this property). If $n$ is such a number, what does this mean about $|\sin n|$?

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    Just inifinitely many won't be enough (e.g. if that happened precisely when $n$ is a perfect square) – 2017-02-07
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    Yes, true, you do actually need that the differences between $n$ that work are bounded. – 2017-02-07