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This question already has an answer here:

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.

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marked as duplicate by Winther, user91500, Watson, Community Aug 20 '16 at 9:20

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    $\begingroup$ You could argue as I do here. $\endgroup$ – David Mitra Aug 20 '16 at 8:03
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    $\begingroup$ if $\sum_{n=1}^{\infty} \frac{\sin^2 n}{n}$ diverges than $\sum_{n=1}^{\infty} \frac{|\sin n|}{n}$ diverges (comparison test). At least one of $\sum_{n=1}^{\infty} \frac{\sin^2 n}{n}$ and $\sum_{n=1}^{\infty} \frac{\cos^2 n}{n}$ diverges, because their sum diverges. $\endgroup$ – miracle173 Aug 20 '16 at 8:04
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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.

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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 $+\infty$.

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