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I am studying limit points of a sequence now, and have some misunderstandings. Here's an exercises I have: The sequence $$\sin \left({n\pi}\over 6\right)$$ has the superior limit $L=1$and the inferior one $l=-1$. Then it is written that: the sequence has 7 limit points:$$l_1=-1,\space l_2={1\over2},\space l_3=-{\sqrt3\over 2},\dots \space l_7=1$$ What does this limit points represent? Thank you very much.

UserX
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wonderingdev
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    i suggest you to edit this to: $\sin(\frac{n\pi}6)$ – mookid Oct 20 '14 at 10:17
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    I have to admit I laughed when I saw \cdot \cdot \cdot – UserX Oct 20 '14 at 10:24
  • I'm perplexed, how is it that $$\overline{\lim_{n\rightarrow \infty}}\left{\sin\left(\frac{n\pi}{6}\right)\right}$$ has 7 limit points? – bjd2385 Oct 20 '14 at 10:27
  • @bd1251252 Period 12 and 5 values achieved twice in each period make for 12-5=7 distinct values. – Did Oct 20 '14 at 10:30
  • Distinct values representing what? – bjd2385 Oct 20 '14 at 10:31
  • @bd1251252 ?? Know what a periodic sequence is? – Did Oct 20 '14 at 10:34
  • You might be talking sideways here. @bd1251252 the question is about the limit points of the original sequence not its limes superior. – quid Oct 20 '14 at 10:43
  • @Did I know what a periodic sequence is, I'm just wondering where the 12 and 5 are coming from. yes, I'm totally ignorant on the subject of superior and inferior limits, I've had the wikipedia crash-course before though. – bjd2385 Oct 20 '14 at 10:47
  • @bd1251252 My previous comment deals only with the number of distinct values and does not touch on limsup and liminf. If you want to understand where 12 and 5 are coming from, posting cryptic comments is not the best way to proceed. Re 12: what might be the period of the sequence of general term sin(n.pi/6), according to you? – Did Oct 20 '14 at 10:52
  • @Did I apologize for my earlier comment(s), just learning. I'm not sure what the period of $$\sin\left(\frac{n\pi}{6}\right)$$ would be...I would assume $\frac{2\pi}{\left|\frac{\pi}{6}\right|}=12$, so there's the 12...what about the 5? – bjd2385 Oct 20 '14 at 10:58
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    @bd1251252 Re the period, note that $\sin((n+12)\pi/6)=\sin((n\pi/6)+2\pi)=\sin(n\pi/6)$. Re the 5, see my answer. – Did Oct 20 '14 at 11:00
  • @Did Okay I see, thanks! – bjd2385 Oct 20 '14 at 11:04

1 Answers1

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Let $x_n=\sin(n\pi/6)$.

The question in the title asks for the limsup of the sequence $(x_n)$, this is solved by noting that if, for every $n$, $x_n\leqslant1$ and, for every $n$, $x_{12n+3}=1$, then $\limsup\limits_nx_n=1$.

The question in the body asks why the sequence $(x_n)$ has $7$ limit points, this is solved by noticing that it has period $12$ and it achieves $5$ values twice in each period, hence there are $12-5=7$ distinct values.

Did
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  • Sorry for a probably stupid question, but can you give me, please, an example? – wonderingdev Oct 20 '14 at 10:36
  • Example of what? – Did Oct 20 '14 at 10:38
  • You said that period 12 and 5 achieved twice in each period make for 12-5=7 distinct values, what do you mean ? Sorry but I dont understand. – wonderingdev Oct 20 '14 at 10:40
  • The period is 12, a priori this makes for 12 numbers. But amongst these, 5 values are achieved twice and 2 values are achieved only once. Thus 5+2 distinct values. – Did Oct 20 '14 at 10:49
  • Alternatively, compute explicitely the first values of the sequence and observe what you get... – Did Oct 20 '14 at 10:54