How can I prove that $\sum_{n=1}^{\infty} \frac{|\sin n|}{n}$ and $\sum_{n=1}^{\infty} \frac{\sin^2 n}{n}$ both diverge?
I thought of using Comparison test, but I couldn't find any sequence to compare with.
This question is from the book 'Real Analysis and Foundations' by Steven G. Krantz.
Let we start from the second one, since it is easier to deal with.
Due to $\sin^2(n) = \frac{1-\cos(2n)}{2}$ we have
$$\sum_{n=1}^{N}\frac{\sin^2 n}{n} = \color{red}{\frac{H_N}{2}}-\sum_{n=1}^{N}
\frac{\cos(2n)}{2n} \tag{1}$$
and the red term ensures (logarithmic) divergence, since $\sum_{n\geq 1}\frac{\cos(2n)}{2n}$ is convergent by Dirichlet's test ($\sum_{n=1}^{N}\cos(2n)$ is bounded). In a similar fashion we may expand $\left|\sin x\right|$ as a Fourier cosine series to get
$$ \left|\sin n\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{m\geq 1}\frac{\cos(2mn)}{4m^2-1}\tag{2} $$
and deduce that
$$ \sum_{n\geq 1}\frac{\frac{2}{\pi}-\left|\sin n\right|}{n} \tag{3}$$
is convergent by Dirichlet's test, so $\sum_{n=1}^N\frac{\left|\sin n\right|}{n}$ diverges like $\frac{2}{\pi}\log N$.
If the asymptotic does not matter, we may just notice that $\left|\sin n\right|\geq\sin^2 n$ as pointed out by miracle173 in the comments.
Hint: You may show that $\sin^2(2n)+\sin^2(2n-1) \geq k=2\sin^2(1/2)>0$ so that the second sum is bounded from below by $\sum_{n\geq 1} \frac{k}{2n}=+\infty$. The first sum is bounded from below by the second (since $|\sin(n)| \geq \sin^2(n)$, hence also equals $+\infty$.
