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I really do not how to attack this problem.

Yung Citgo
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    A way to do it is to use $|\sin (n) |\ge \sin^2(n) = \frac{1}{2} - \frac{1}{2} \cos(2n)$. Using the divergence of $\sum \frac{1}{n}$ and the convergence of $\sum \frac{\cos(2n)}{n}$, you can conclude – charmd Oct 27 '16 at 17:18
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    Another way is to note that among any four consecutive integers, there is at least one with $\lvert \sin n\rvert \geqslant 1/2$. Thus you have a lower bound for the partial sums in terms of harmonic numbers. – Daniel Fischer Oct 27 '16 at 17:22
  • I added an answer based on Daniel Fischer comment in the link marked as duplicated. – Darío G Oct 27 '16 at 17:58

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