Question about $\sum_{n=1}^{\infty}\frac{|\sin(n)|}{n}$. In several places on this site the sum $\displaystyle\sum_{n=1}^{\infty}\dfrac{\sin(n)}{n}$ has been discussed as a generalized alternating series, which therefore converges. I am curious about the sum $\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}$, which I am sure does not converge.  Here is my "proof".  Steps 2 and 3 clearly needs some clarification.
Step 1) Define $A\subset\mathbb{R}$ by $A:=[\pi/4,3\pi/4]\cup[5\pi/4,7\pi/4]\cup\ldots$.  For any $x\in A$, $|\sin(x)|\geq\sqrt{2}/2$.
Step 2) Since $\pi$ is irrational, roughly half the integers are in $A$.  That is, if we define $\#_A(n)$ to be the cardinality of $\{1,\ldots,n\}\cap A$, $\displaystyle\lim_{n\to\infty}\dfrac{\#_A(n)}{n}=\dfrac{1}{2}$.
Step 3) Therefore $\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}>\sum_{n>0,n\in A}\dfrac{|\sin(n)|}{n}>\sum_{n>0,n\in A}\dfrac{\sqrt{2}/2}{n}\approx\dfrac{1}{2}\sum_{i=1}^n\dfrac{\sqrt{2}/2}{n}$.
Step 4) Since the last sum in Step 3 diverges, the sum $\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}$ diverges.
I beleive this is more or less intuitively right, but I do not really know the theory to make rigorous Steps 2 and 3.
 A: Your idea is almost correct, except that we need some justification. So here is the missing part:

Lemma. If $(a_{n})_{n=1}^{\infty}$ be such that $\sigma_{n} = (a_{1}+\cdots+a_{n})/n$ converges to $\alpha > 0$, then
  $$ \sum_{n=1}^{\infty} \frac{a_{n}}{n} = \infty. \tag{1}$$

Proof. Let $s_{n} = a_{1} + \cdots + a_{n}$ denotes the partial sum. Then for $n \geq 2$,
$$ \frac{a_{n}}{n} = \sigma_{n} - \sigma_{n-1} + \frac{\sigma_{n-1}}{n}. $$
Consequently, we have
$$ \sum_{n=1}^{N} \frac{a_{n}}{n} = \sigma_{N} + \sum_{n=1}^{N-1} \frac{\sigma_{n}}{n+1}. $$
Using standard techniques, it is easy to see that
$$ \frac{1}{\log N} \sum_{n=1}^{N-1} \frac{\sigma_{n}}{n+1} \xrightarrow[]{N\to\infty} \alpha. $$
Thus we have
$$ \sum_{n=1}^{N} \frac{a_{n}}{n} = (\alpha+o(1))\log N $$
and hence (1) follows. ////
A: $f(x)=x;- \pi \le x \le \pi$
$x \in \left[ { - \pi ;\pi } \right]; - x \in \left[ { - \pi ;\pi } \right]:f( - x) =  - f(x)$
$\forall n \in N:a_n= 0$
$b_n  = \frac{2}{\pi }\int\limits_0^\pi  {x\sin nxdx = }  - \frac{2}{{\pi n}}\left[ {x\cos nx} \right]_0^\pi$
$ b_n  =  - \frac{2}{{n\pi }}\left[ {x\cos nx} \right]_0^\pi   = \frac{2}{n}\left( { - 1} \right)^{n + 1}$
$f(x) = 2\sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} \sin nx}}{n}}  $
$x\in\left] { - \pi ;\pi } \right]:\frac{x}{2} = \sum\limits_{n = 1}^{ + \infty } {\left( { - 1} \right)^{n + 1} \frac{{\sin nx}}{n}}$
$ t \in \left] {0;2\pi } \right]:\left( {\pi  - t} \right) \in \left] { - \pi ;\pi } \right]$
if pose  $x = \pi  - t$
$\displaylines{
  \sin \left( {n\left( {\pi  - t} \right)} \right) = \cos n\pi \sin \left( { - nt} \right) + \sin n\pi \cos \left( { - nt} \right) \cr 
   = \left( { - 1} \right)^n \sin \left( { - nt} \right) = \left( { - 1} \right)^{n + 1} \sin \left( {nt} \right) \cr}$
$\sum\limits_{n = 1}^{ + \infty } {\frac{{\sin nt}}{n}}  = \frac{{\pi  - t}}{2};\forall t \in \left] {0;2\pi } \right]$
t=1
$\sum\limits_{n = 1}^{ + \infty } {\frac{{\sin n}}{n}}  = \frac{{\pi  - 1}}{2}$
A: You could use the fact that the intervals $\left(\frac{(4n-1)\pi}{4},\frac{(4n+1)\pi}{4}\right)$ have length $\frac{\pi}{2}<2$, 
so there can't be 3 consecutive integers in any of these intervals;
so $S_{3n}=\frac{|\sin 1|}{1}+\frac{|\sin 2|}{2}+\frac{|\sin3|}{3}+\cdots+\frac{|\sin 3n|}{3n}\ge\frac{\sqrt{2}}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\cdots+\frac{1}{3n}\right)\to\infty$.

Another argument you could use  is that
$\displaystyle\frac{|\sin n|}{n}\ge\frac{\sin^2 n}{n}=\frac{\frac{1}{2}(1-\cos 2n)}{n}\ge0$ for all n, and
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ diverges while $\displaystyle\sum_{n=1}^{\infty}\frac{\cos2n}{n}$ converges by Dirichlet's test,
so $\displaystyle\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$ diverges by the Comparison Test.
