I'm not sure whether the following series converges or diverges: $$\sum _{n=1}^{\infty }\frac{\left|\sin (n)\right|}{n}$$

I proved that the series $\sum _{n=1}^{\infty }\frac{\sin^2 (n )}{n}$ converge. Is there a way I can use that? I've tried using Dirichlet series test with the latter but didn't got nowhere since $\frac{1}{\left|\sin x\right|}$ is not monotone decreasing.

  • 1
    $\begingroup$ The series you are claiming to converge diverges. We can use this to prove that the series in question diverges. $\endgroup$
    – Jakobian
    Jun 2, 2018 at 15:05
  • 3
    $\begingroup$ $|\sin n|>1/2$ often enough to ensure that $\sum|\sin n|/n$ diverges. $\endgroup$ Jun 2, 2018 at 15:09

3 Answers 3


$$ \frac{\sin^2(n)}{n} = \frac{1}{2n} - \frac{\cos(2n)}{2n} $$ By Dirichlet's test, $\sum \frac{\cos(2n)}{2n} $ converges, hence $\sum \frac{\sin^2(n)}{n} $ diverges (since $\sum \frac{1}{n}$ diverges to $\infty$). $$ \frac{\sin^2(n)}{n} \leq \frac{|\sin(n)|}{n}$$ So by the comparision test, $\sum \frac{|\sin(n)|}{n} $ diverges

  • $\begingroup$ It isn't obvious to me that $\sum_{n=1}^N{\cos(2n)}$ is uniformly bounded. How does one see this? $\endgroup$
    – saulspatz
    Jun 2, 2018 at 15:32
  • $\begingroup$ Sum of sines or cosines $\endgroup$
    – Jakobian
    Jun 2, 2018 at 15:36

Another way to do this: Consider $e^{in}, n = 1,2,\dots, 7$ One of those points lies in the arc from $\pi/6$ to $\pi/2.$ For that $n,$ we have $(\sin n)/n \ge (1/2)/ n.$ Thus we can say

$$\sum_{n=1}^{7}\frac{|\sin n|}{n} \ge \frac{1/2}{7 }.$$

The same idea works for the next $7$ integers, so

$$\sum_{n=8}^{14}\frac{|\sin n|}{n} \ge \frac{1/2}{14 }.$$

And so on.. We can conclude

$$\sum_{n=1}^{\infty}\frac{|\sin n|}{n} \ge \sum_{k=1}^\infty\frac{1/2}{7k }=\infty.$$

Now why $7$ and why that particular arc? Easy: $7$ is the first integer $>2\pi,$ and that arc has length $>1,$ forcing $e^{in}$ to land in there at least once every orbit.

  • $\begingroup$ How do you prove one of every $7$ points lies in the arc from $\pi/6$ to $\pi/2$? $\endgroup$
    – Hans
    Dec 13, 2019 at 18:52
  • $\begingroup$ @Hans Did you read the last paragraph? $\endgroup$
    – zhw.
    Dec 13, 2019 at 19:19
  • $\begingroup$ Oops, I did not read to the last paragraph. Let us delete these comments. Nice solution. +1 $\endgroup$
    – Hans
    Dec 14, 2019 at 8:12

Note that, since $\pi/3\gt1\gt\pi/4$, we have $\sin1\gt\cos1\gt\cos(\pi/3)=1/2$. It follows that

$$\sin^2(n-1)+\sin^2(n+1)=(\sin n\cos1-\cos n\sin 1)^2+(\sin n\cos1+\cos n\sin1)^2\\ =2(\sin^2n\cos^21+\cos^2n\sin^21)\\ \gt2(\sin^2n+\cos^2n)/4=1/2$$

Therefore, taking three consecutive terms of the sum, we see that


(by ignoring the middle term altogether, using the general inequality $|\sin x|\ge\sin^2x$, and taking the larger of the two remaining denominators). Thus

$$\sum_{n=1}^\infty{|\sin n|\over n}\gt|\sin1|+\sum_{k=1}^\infty{1\over6k+2}=\infty$$

(Note, there was no real need to include the first term, $|\sin1|$ on the right hand side in the final inequality; I kept it just only to point out that it did not participate in any of the triples of consecutive terms, which were all centered on multiples of $3$.)


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