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?
 A: HINT Given any $4$ consecutive numbers, then at-least one of them will have $\vert \sin(n) \vert > 1/2$.
A: It is not convergent:
$$
\frac{|\sin n|}{n}\ge\frac{\sin^2n}{n}=\frac12\Bigl(\frac1n-\frac{\cos(2\,n)}{n}\Bigr).
$$
A: 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.
