Determine if $\displaystyle\sum_{k=1}^\infty\sin(kx)\sin\left(\cfrac{1}{kx}\right)$ is convergent Determine if the following series converges absolutely:
$$\sum_{k=1}^\infty \sin(kx)$$
and
$$\sum_{k=1}^\infty\sin(kx)\sin\left(\cfrac{1}{kx}\right)$$
I know how to deal with whether they converge. First one diverge by n-term test, second one converge by Dirichlet's Test. However, I do not know how to deal with whether they converge absolutely. Can someone teach me how to determine if both series converge absolutely. Thanks.

The series 
$$\tag1 \sum_{k=1}^\infty \sin(kx)$$
converges absolutely for $x\in\pi \mathbb Z$. For all other cases, it doesn' even converge: 
Note that
$$ \sin((k+1)x)=\sin(kx)\cos x+\cos(kx)\sin x$$
hence whenever $\sin(kx)\approx 0$ and hence $|\cos(kx)|\approx1$, then $|\sin((k+1)x)|\approx|\sin x|$. 
More precisely, if $\sin x\ne0$, we can find $a>0$ with
$$ \sqrt{1-a^2}\cdot |\sin x|>2a.$$
Then $|\sin(kx)|<a$ implies $|\cos(kx)|>\sqrt{1-a^2}$ and hence
$$ |\sin((k+1)x)\ge |\cos(kx)\sin x|-|\sin(kx)\cos x|> \sqrt{1-a^2}|\sin x|-a>a$$
We conclude that $\sin(kx)\not\to 0$ unless $\sin x=0$.
In summary, the series $(1)$


*

*converges absolutely (trivially) for $x\in\pi\mathbb Z$

*diverges for $x\notin\pi\mathbb Z$.



We have seen above that for $x\notin\pi\mathbb Z$, at least one of two successive summands is $>a$ in absolute value.
Apply this to
$$\tag2 \sum_{k=1}^\infty \left|\sin(kx)\sin(\tfrac1{kx})\right|$$
For $k$ sufficiently large we have $\sin\frac1{kx}\approx\frac1{kx}$, say $|\sin\frac1{kx}|\ge\frac1{2k|x|}$.
Then for such $k$ we have $$\left|\sin(kx)\sin(\tfrac1{kx})\right|+\left|\sin((k+1)x)\sin(\tfrac1{(k+1)x})\right| \ge \frac{a}{2(k+1)|x|}.$$
We conclude by comparision with the harmonic series that $(2)$ diverges  for $x\notin\pi\mathbb Z$. In summary, the series
$$ \sum_{k=1}^\infty\sin(kx)\sin\frac1{kx}$$


*

*is not defined for $x=0$

*converges absolutely (trivially) for $x\in\pi\mathbb Z\setminus\{0\}$

*converges (as you have shown, by the Dirichlet test), but not absolutely, for $x\in\mathbb R\setminus\pi\mathbb Z$.

