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Let $ u_{n} = \sin \! \left( \dfrac{\pi}{n} \right) $, where $ n \in \Bbb{N} $, and consider the series $ \displaystyle \sum_{n = 1}^{\infty} u_{n} $. Which of the following is/are true?

(a) $ \displaystyle \sum_{n = 1}^{\infty} u_{n} $ is convergent.

(b) $ \displaystyle \sum_{n = 1}^{\infty} u_{n} $ is divergent.

(c) $ \displaystyle \sum_{n = 1}^{\infty} u_{n} $ is absolutely convergent.

(d) $ u_{n} \to 0 $ as $ n \to \infty $.

Now, $ n \to \infty $ implies $ \dfrac{\pi}{n} \to 0 $, so $ u_{n} = \sin \! \left( \dfrac{\pi}{n} \right) \to 0 $. Also, from the graph of $ \sin $, it looks like this sequence will tend to $ 0 $.

I am not sure about the series options — whether they are all wrong or some are right, and why so.

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  • $\begingroup$ option $b)$ is true. $\endgroup$ – DeepSea Aug 29 '15 at 7:19
  • $\begingroup$ @Ganymede : is $d$) wrong ? $\endgroup$ – user118494 Aug 29 '15 at 7:21
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    $\begingroup$ No....(d) is correct...(b) is also correct... $\endgroup$ – Empty Aug 29 '15 at 7:22
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For $x\le\frac\pi2$, concavity implies $\frac2\pi x\le\sin(x)\le x$.

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Therefore, $$ \begin{align} \sum_{n=1}^\infty\sin\left(\frac\pi n\right) &=\sum_{n=2}^\infty\sin\left(\frac\pi n\right)\\ &\ge\frac2\pi\sum_{n=2}^\infty\frac\pi n\\ &=2\sum_{n=2}^\infty\frac1n \end{align} $$ which diverges.

Furthermore, $$ \lim_{n\to\infty}\sin\left(\frac\pi n\right)\le\lim_{n\to\infty}\frac\pi n=0 $$

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$sin(\frac{\pi}{n})$ is asymptotically equivalent to $\frac{\pi}{n}$ so it behaves like the harmonic series which is divergent.

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  • $\begingroup$ "equivalent to"? $\endgroup$ – 6005 Aug 29 '15 at 7:37
  • $\begingroup$ en.wikipedia.org/wiki/Asymptotic_analysis $\endgroup$ – curiosity Aug 29 '15 at 7:39
  • $\begingroup$ I though that the word "equivalent" in a context like this one has a canonical meaning. Are you ok with "asymptotically equivalent" ? $\endgroup$ – curiosity Aug 29 '15 at 7:47
  • $\begingroup$ yes, I like that wording better. I think this was a minor point anyway. $\endgroup$ – 6005 Aug 29 '15 at 7:51
  • $\begingroup$ As long as we know what we're talking about the rest is just vocab ... anyway I've updated my answer $\endgroup$ – curiosity Aug 29 '15 at 7:55
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Hint :

Take , $v_n=\frac{1}{n}$. Then use comparison test.

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Since $\sin \frac{\pi}{n} \sim \frac{\pi}{n}$ as $n \to \infty$, and since $\sum_{n \geq 1}\frac{\pi}{n}$ diverges, by the limit comparison test we see that $\sum_{n\geq 1}\sin \frac{\pi}{n}$ diverges.

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