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By using the integral test, I know that $$\sum\limits_{n=1}^\infty \sin\left(\frac{1}{n}\right)$$ diverges. However, how would I show that the series diverges using the limit comparison test? Would I simply let $\sum\limits_{n=1}^\infty a_n = \sum\limits_{n=1}^\infty b_n = \sum\limits_{n=1}^\infty \sin\left(\frac{1}{n}\right)$ and then take $\displaystyle \lim_{n \rightarrow \infty}\frac{a_n}{b_n}$ to show the series diverges (assuming the limit converges to some nonnegative, finite value)?

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  • $\begingroup$ For Limit Comparison, compare with $\sum \frac{1}{n}$. $\endgroup$ Feb 4, 2014 at 0:47
  • $\begingroup$ Why not do straight comparison with $\sum \frac{1}{n}$? $\endgroup$ Feb 4, 2014 at 1:10
  • $\begingroup$ Disregard my comment. What I was thinking was nonsense. $\endgroup$ Feb 4, 2014 at 1:17

2 Answers 2

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Notice $x_n = \frac{1}{n} $, then $\sum \frac{1}{n}$, the harmonic series, we all know is divergent. Now,

$$ \lim \frac{ \sin (\frac{1}{n})}{\frac{1}{n}} =_{t = \frac{1}{n}} \lim_{t \to 0} \frac{ \sin t}{t} = 1$$

The result now follows by the limit comparison test. :)

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Note the following for: For $x\in [0, \frac{\pi}{2}]$ we have, $$\color{blue}{\frac{2x}{\pi}\le\sin x\le x}$$

Taking $x =\frac{1}{n}$ we get,

$$\sum\limits_{n=1}^\infty \sin\left(\frac{1}{n}\right) \ge \frac{2}{\pi}\sum\limits_{n=1}^\infty \frac{1}{n} =\infty$$

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