Does $\sum\limits_{k=1}^\infty |\sin(ak)/k|$ converge? Does $\sum_{k=1}^\infty |\sin(ak)/k|$ converge for all $0<a<\pi$? I do not think so since for $a=\pi
/2$:
$$\sum_{k=1}^\infty\left\vert\frac{\sin(ak)}{k}\right\vert=\sum_{k=0}^\infty (2k+1)^{-1}=\infty.$$
But is this true for all $a$?
 A: 
The series diverges if and only if $\sin(a)\ne0$, that is, if and only if $a/\pi$ is not an integer.

For every $k$, let $N_k^\varepsilon=\{n\leqslant k\mid|\sin(an)|\geqslant\varepsilon\}$, then the sum of the series is at least $$\sum_{k\geqslant1}\frac{\varepsilon}k\mathbf 1_{|\sin(ak)|\geqslant\varepsilon}=\varepsilon\sum_{k\geqslant1}\frac1k(N_k^\varepsilon-N_{k-1}^\varepsilon)=\varepsilon\sum_{k\geqslant1}\frac{N_k^\varepsilon}{k(k+1)},$$ hence it suffices to show that, for some positive $\varepsilon$, there exists some positive $\alpha$ and some finite $n$ such that, for every $k\geqslant n$, $$N_k^\varepsilon\geqslant\alpha k,$$ to conclude that the series diverges.
Assume without loss of generality that $0\leqslant a\leqslant2\pi$. If $a=0$ or $a=\pi$ or $a=2\pi$, then every term is zero hence the series converges. Otherwise, let us show that two successive terms $|\sin(ak+a)|$ and $|\sin(ak)|$ cannot both be small. 
To wit, assume that $|\sin(ak)|\leqslant\varepsilon$ for some $\varepsilon$ in $(0,1)$, then $$\sin(ak+a)=\sin(ak)\cos(a)+\cos(ak)\sin(a),$$ and $\sin(a)\ne0$ hence $$|\sin(ak+a)|\geqslant|\cos(ak)|\cdot|\sin(a)|-|\sin(ak)|\cdot|\cos(a)|\geqslant\sqrt{1-\varepsilon^2}\cdot|\sin(a)|-\varepsilon.$$ For every $\varepsilon\leqslant\frac13|\sin(a)|$, the RHS is at least $\varepsilon$. For such value of $\varepsilon$, this proves our claim that $|\sin(ak+a)|$ and $|\sin(ak)|$ cannot both be less than $\varepsilon$, hence $N_k^\varepsilon$ is at least of order $\frac12k$ and the result follows.
