3

Given is a sequence with:

$(a_0)=1$, $(a_1=1)$, $a_{n+2}=\frac{1+a_{n+1}}{a_n}$

I now have to show what the accumulation points are:

I guess that the sequence is jumping from number to number like this: 1->1->2->3->2->1->1->2..

So the acc.points should indeed be "1, 2 and 3".

Is this correct? If yes, how can I 'show' this? Furthermore: Is there any way to build some subsequence of $a_n$ that converges against "1, 2 and 3" ?

Thank you :)

Vazrael
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    Mhm thought about: every sequence of type $a_{5x}$ converges against (or should I say: "to"?) 1. $a_{5x +1}$ also does. $a_{5x +2}$ converges against 2 and $a_{5x+3}$ against 3? Is this sufficient? – Vazrael May 13 '13 at 13:06
  • You can prove it rigorously by induction. – Pedro May 13 '13 at 13:09
  • Kevin, I have changed my answer after some discussion. – Pedro May 13 '13 at 18:17
  • Note, that the sequence gives the periodic decimal expansion of the fraction 1248/11111. – DVD May 13 '13 at 23:02

1 Answers1

3

Prove by induction that for $n=4,9,14,\dots$, $a_n=3$, that for $n=0,1,5,6,10,11,\dots$, $a_n=1$ and that for the remaining indices, $a_n=2$. That is all you want to do.

As per the accumulation points, recall that a number $\alpha$ is an accumulation point or limit point of a sequence $\langle a_n\rangle$ if there exists a subsequence of $\langle a_n\rangle$ that converges to $\alpha$. Thus, $1,2,3$ are accumulation points of $\langle a_n\rangle$.

Pedro
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  • So there aren't any accumulation points of $a_n$ at all? – Vazrael May 13 '13 at 13:26
  • @KevinLiebing I have changed my answer. – Pedro May 13 '13 at 18:32
  • @ Peter Tamaroff: I'm glad you posted it in the first way. Since I'm not a native english speaker, the words for "accumulation" and "limit" point are indeed hot-discussed; some do make a difference between them, therefore 1-2-3 wouldn't be limit-points (since they refer to limit-point being related to a set), however 1-2-3 are accumulation-points.. Anyway, thank you! – Vazrael May 15 '13 at 06:25