# 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 first converges only if $x\in\pi\mathbb Z$, and then it also converges absolutely. – Hagen von Eitzen Apr 28 '14 at 15:46
• what about other values of x? – user10024395 Apr 28 '14 at 15:51

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 $x\in\mathbb R\setminus\pi\mathbb Z$.
• +1 Nice argument, self-contained and does not require equidistribution theorem, unlike the one I proposed. – Marcin Łoś Apr 28 '14 at 16:11
• i don't really understand this part: The "≈" of above can be made precise i nthe followin sense: Unless sinx=0, there exists a positive number a>0 such that |sin(kx)|<a implies |sin((k+1)x)>a. In other words: at least one of two successive summands is >a in absolute value. – user10024395 Apr 28 '14 at 16:25
• can u explain more pls thanks – user10024395 Apr 28 '14 at 16:26
• i am really sorry for asking you another time, but even after the edit, i still don't understand why we can find such a – user10024395 Apr 28 '14 at 23:56

The first one, as it diverges, cannot converge absolutely (except for trivial case $x\in \pi\mathbb{Z}$). As for the second one - it is not absolutely convergent for most values of $x$ (when $x$ is not a rational multpile of $\pi$), the reason being roughly that in such case "many" values of $sin\ kx$ are "quite large", greater in absolute value than some fixed $0<a<1$. As the second term - $\sin \frac{1}{kx}$ - can be approximated by $\frac{1}{kx}$, which is divergent, it follows that partial sums are unbounded. Of course, it's just a vague sketch of a proof. You can find precise formulation of the above ideas and relevant results here