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How to prove that $ \sum_{n \in \mathbb{N} } | \frac{\sin( n)}{n} | $ diverges?

Does the series $\displaystyle\sum_{n=1}^{\infty} \frac{|\sin(n)|}{n}$ converge or diverge? (And why?)

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Sorry guys...I did look believe it or not! –  Freddie Jan 14 '13 at 22:21
    
This question also covers this result. –  robjohn Jan 14 '13 at 22:22
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marked as duplicate by mrf, amWhy, Marvis, P.., Ross Millikan Jan 14 '13 at 22:34

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Note that $|\sin{x}| \geqslant {\dfrac{1}{2}}$ for $x \in I_k=\left[{\dfrac{\pi}{6}}+k\pi, \;\;\pi-{\dfrac{\pi}{6}}+k\pi \right].$ Length of every $I_k$ $$|I_k|=\pi-{\dfrac{2\pi}{6}}={\dfrac{2\pi}{3}}>2,$$ so every $I_k$ contains at least one natural number $n_k.$

Then $$\sum\limits_{n=1}^{N} \dfrac{|\sin(n)|}{n} \geqslant {\dfrac{1}{2}}\sum\limits_{n_k\leqslant{N}} \dfrac{1}{n_k} \underset{N\to\infty}{\to}{\infty}$$ since the harmonic series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n}$ diverges.

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This is not correct. You need $1/n_k$ in your sum, and there is no easy way to show that the "thinned out" version of the harmonic series diverges. (What happens if $n_k = 2^k$?) –  mrf Jan 14 '13 at 22:39
    
@mrf $|\sin{x}| \geqslant {\dfrac{1}{2}}$ for $x \in I_k=\left[{\dfrac{\pi}{6}}+k\pi, \;\;\pi-{\dfrac{\pi}{6}}+k\pi \right].$ Length of every $I_k$ $$|I_k|=\pi-{\dfrac{2\pi}{6}}={\dfrac{2\pi}{3}}>2,$$ so, every $I_k$ contains at least one natural number $n_k.$ –  M. Strochyk Jan 14 '13 at 22:57
    
Yes, but that is quite far from what you wrote first. –  mrf Jan 14 '13 at 22:59
    
@mrf Thanks for pointing of my inaccuracy –  M. Strochyk Jan 14 '13 at 23:04
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