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It is stated as a problem in Spivak's Calculus and I can't wrap my head around it.

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The most promising thing I've tried is picking n so that $| \sin n | > x$, where x is some number between 0 and 1. Then we have that my original series is larger than something SIMILAR to the harmonic series. To be exact, harmonic series without those n where $| \sin n | \leq x$. Intuitively, we should be able to pick x so that those n are ''rare enough'', but I can't prove it rigorously. – ante.ceperic Dec 25 '12 at 16:03
Hint: for each $k\in\Bbb N$, the interval $[k\pi+{\pi\over 6},(k+1)\pi-{\pi\over 6}]$ contains an integer. Using this and the comparison test, show that some subseries of your series diverges. – David Mitra Dec 25 '12 at 16:07
up vote 15 down vote accepted

Hint: $|\sin n|\geqslant \sin^2 n$, and use the convergence of $\sum_{n=1}^{+\infty}\frac{\cos(\color{red}2n)}n$ and the divergence of harmonic series.

We have $\cos(2n)=2\cos^2n-1$, and $$\sin^2n=1-\cos^2n=1-\frac{\cos(2n)+1}2=\frac{1-\cos(2n)}2.$$

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The idea in your comment leads to one approach:

For each positive integer $k$, the interval $\bigl[k\pi+{\pi\over 6}, (k+1)\pi-{\pi\over 6}\bigr]$ has length exceeding $1$ and thus contains an integer $n_k$. We then have ${|\sin (n_k)|\over n_k} \ge {\sin({\pi\over 6})\over (k+1)\pi}$. So, from the Comparison test, it follows that $\sum\limits_{k=1}^\infty {|\sin(n_k)|\over n_k}$ diverges; whence $\sum\limits_{n=1}^\infty {|\sin(n )|\over n}$ diverges (by the Comparison test again).

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Thank you very much for this! – ante.ceperic Dec 25 '12 at 16:37
@David Mitra: how did you choose this specific interval($[\pi k +\frac{\pi}{6},\pi(k+1)-\frac{\pi}{6}$? – Alex Jun 23 '13 at 2:38

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