# Does $\sum _{n=1}^{\infty }\frac{\left|\sin\left(n\right)\right|}{n}$ converge?

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.

• The series you are claiming to converge diverges. We can use this to prove that the series in question diverges. Jun 2 '18 at 15:05
• $|\sin n|>1/2$ often enough to ensure that $\sum|\sin n|/n$ diverges. Jun 2 '18 at 15:09

$$\frac{\sin^2(n)}{n} = \frac{1}{2n} - \frac{\cos(2n)}{2n}$$ 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

• It isn't obvious to me that $\sum_{n=1}^N{\cos(2n)}$ is uniformly bounded. How does one see this? Jun 2 '18 at 15:32
• Sum of sines or cosines Jun 2 '18 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.

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

$${|\sin(3k-1)|\over3k-1}+{|\sin3k|\over3k}+{|\sin(3k+1)|\over3k+1}\gt{\sin^2(3k-1)+\sin^2(3k+1)\over3k+1}\gt{1\over2(3k+1)}$$

(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 $3$.)