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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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Hi and welcome to the site. Both you and the site would benefit if you would consider registering your account. Also, if you find answers helpful, up vote them. If you find a "best answer", accept it by clicking on the check mark under the question. – JohnD Jan 14 '13 at 22:18
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Sorry guys...I did look believe it or not! – Freddie Jan 14 '13 at 22:21
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This question also covers this result. – robjohn Jan 14 '13 at 22:22
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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.
M. Strochyk
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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
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@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
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