How to prove that $ \sum_{n \in \mathbb{N} } | \frac{\sin( n)}{n} | $ diverges? It is stated as a problem in Spivak's Calculus and I can't wrap my head around it. 
 A: Hint: $|\sin n|\geqslant \sin^2 n$, and use the convergence of $\sum_{n=1}^{+\infty}\frac{\cos(\color{red}2n)}n$ and the divergence of harmonic series. 
We have $\cos(2n)=2\cos^2n-1$, and 
$$\sin^2n=1-\cos^2n=1-\frac{\cos(2n)+1}2=\frac{1-\cos(2n)}2.$$
A: The idea in your comment leads to one approach:
For each positive integer $k$, the interval $\bigl[k\pi+{\pi\over 6}, (k+1)\pi-{\pi\over 6}\bigr]$ has length exceeding $1$ and thus contains an integer $n_k$. We then have ${|\sin (n_k)|\over n_k} \ge {\sin({\pi\over 6})\over (k+1)\pi}$.
So, from the Comparison test, it follows that $\sum\limits_{k=1}^\infty  {|\sin(n_k)|\over n_k}$ diverges; whence $\sum\limits_{n=1}^\infty  {|\sin(n )|\over n}$ diverges (by the Comparison test again).
A: A slightly different approach. Assuming that $\sum_{n\geq 1}\frac{\left|\sin n\right|}{n}$ is convergent, Kronecker's lemma ensures that $\{\left|\sin n\right|\}_{n\geq 1}$ has mean zero. Now there are at least a couple of ways  (besides $\left|\sin n\right|\geq \sin^2 n$) for showing this is absurd:


*

*By the equidistribution theorem, $e^{in}$ essentially is a random point on $S^1$ for $n\in\mathbb{N}$, hence $\left|\sin n\right|$ essentially behaves like a random variable, whose density is supported on $(0,1)$ and given by $\frac{2}{\pi\sqrt{1-x^2}}$;

*The Fourier cosine series of $\left|\sin x\right|$ is given by
$$ \left|\sin x\right|=\color{red}{\frac{2}{\pi}}-\frac{4}{\pi}\sum_{m\geq 1}\frac{\cos(2mx)}{4m^2-1} $$
hence the average value of $\{\left|\sin n\right|\}_{n\geq 1}$ is $\frac{2}{\pi}>0$.
