Prove that the following series do not converge absolutely Prove that the following series do not converge absolutely:
$$\sum_{x=1}^\infty \frac{\sin x}{x}$$
 A: Note that $|\sin x| \geq \sin^2 x$. If $\sum_{x=1}^\infty \frac{|\sin x|}{x} < \infty$, by comparison test we have
$$
\sum_{x=1}^\infty \frac{\sin^2 x}{x} \leq \sum_{x=1}^\infty \frac{|\sin x|}{x} < \infty
$$
However, 
$$
\sum_{x=1}^\infty \frac{\sin^2 x}{x} = \sum_{x=1}^\infty \frac{1 - \cos 2x}{2x}
$$
As $\sum_{x=1}^\infty \frac{1}{2x} = \infty$ and $\sum_{x=1}^\infty \frac{\cos 2x}{2x}$ converges(you may show by Dirichlet's test), $\sum_{x=1}^\infty \frac{1 - \cos 2x}{2x}$ diverges, a contradiction.
A: The idea behind the proof below is that for any two consecutive integers $x$ and $x+1$, at least one of $|\sin(x)|$ and $|\sin(x+1)|$ is "big."
We use the fact that $\sin(x+1)=\sin x\cos 1 +\cos x\sin 1$. Suppose that $|\sin x|\le \frac{1}{3}$. Then $|\cos x|\ge \sqrt{1-\frac{1}{9}}$. 
Thus if $|\sin x\le \frac{1}{3}$ then
$$|\sin(x+1)|\ge \sqrt{1-\frac{1}{9}}\sin 1-\frac{1}{3}\cos 1\approx  0.6.$$
  It follows that in $n$ and $n+1$ are a pair of consecutive integers, then at least one of $|\sin(n)|$ and $|\sin(n+1)$ is greater than $\frac{1}{3}$.  Thus
$$\frac{|\sin(2k+1)|}{2k+1}+\frac{|\sin(2k)|}{2k} \ge \frac{1}{6k}.$$
But $\sum_{k=1}^\infty \frac{1}{6k}$ diverges, so $\sum_{x=1}^\infty \frac{|\sin x|}{x}$ diverges. 
