# Covergence test of $\sum_{n\geq 1}{\frac{|\sin n|}{n}}$

I need to prove that $$\sum_{n\geq 1}{\frac{|\sin n|}{n}}$$ is convergent.

How should I do it?

• All convergence tests I know. – Breldor Nov 22 '14 at 19:52
• The problem is, of course, that that series is divergent. – Daniel Fischer Nov 22 '14 at 19:52

It is not convergent: $$\frac{|\sin n|}{n}\ge\frac{\sin^2n}{n}=\frac12\Bigl(\frac1n-\frac{\cos(2\,n)}{n}\Bigr).$$

• Don't you also need to show that $\sum \frac{\cos 2n}{n}$ is convergent? – TonyK Nov 23 '14 at 17:23
• I am assuming this is known. Use Dirichlet's test to prove it. – Julián Aguirre Nov 23 '14 at 17:29
• Can you replace the last part with $1/n - \cos^2{n} /n$? – Whyka Jun 17 '15 at 21:55
• @Whyka Yes, but then you have to prove that $\sum\cos^2n/n$ does not converge. – Julián Aguirre Jun 18 '15 at 9:27

HINT Given any $4$ consecutive numbers, then at-least one of them will have $\vert \sin(n) \vert > 1/2$.

• We could probably stretch this to $3$ consecutive numbers instead of $4$ consecutive numbers. – Adhvaitha Nov 22 '14 at 20:23
• But everytime |sin n| is being divided by a greater number too, so how to conclude that it is divergent? – kaka Nov 23 '14 at 17:56
• $$\dfrac{\vert \sin(n) \vert}n + \dfrac{\vert \sin(n+1) \vert}{n+1} + \dfrac{\vert \sin(n+2) \vert}{n+2} > \dfrac1{2n}$$ – Adhvaitha Nov 23 '14 at 17:58
• Actually the argument gives only $$\frac{|\sin(n)|}n+\frac{|\sin(n+1)|}{n+1}+\frac{|\sin(n+2)|}{n+2}\gt\frac1{2(n+2)}.$$ – Did Nov 23 '14 at 20:38
• @Did Yes, you are right. – Adhvaitha Nov 23 '14 at 20:40

This is the same idea as in the answer given by Advhaitha, with some more details:

Recall that the period of the function $f(x)=|\sin(x)|$ is $\pi$. Also, we can consider the discrete set $A=\{x:|\sin(x)|=\frac{1}{2}\}=\{x=k\pi \pm \frac{\pi}{6}: k\in\mathbb{Z}\}$ Notice that the difference between two consecutive elements in $A$ is at most $\frac{2\pi}{3}<3$.

So, for every fixed $n$ we have that one of the values $|\sin(n)|,|\sin(n+1)|,|\sin(n+2)|,|\sin(n+3)|$ must be bigger than $\frac{1}{2}$.

So, \begin{align*}\sum_{n=1}^\infty \dfrac{|\sin n|}{n}&\geq \sum_{k=0}^\infty \left(\dfrac{|\sin(4k+1)|}{4k+1}+\dfrac{|\sin(4k+2)|}{4k+2}+\dfrac{|\sin(4k+3)|}{4k+3}+\dfrac{|\sin(4k+4)|}{4k+4}\right)\\ &\geq\sum_{k=0}^\infty \dfrac{1/2}{4k+4}=\dfrac{1}{8}\cdot \sum_{k=0}^\infty \dfrac{1}{k+1}=\dfrac{1}{8}\sum_{n=1}^\infty \dfrac{1}{n}. \end{align*}

Since the last series diverges, we conclude by comparison that the series $\displaystyle{\sum_{n=1}^\infty \dfrac{|\sin n|}{n}}$ diverges too.