Recursive sequence convergence

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Homework Help Overview

The discussion revolves around a recursive sequence defined by s_{1} = 1 and s_{n+1} = \sqrt{s_{n} + 1}. Participants are tasked with proving that this sequence converges to \frac{1}{2}(1+\sqrt{5}).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the recursive nature of the sequence and its convergence properties. Some attempt to establish bounds for the sequence, while others question the assumptions made about the limit. There is discussion about the necessary conditions for convergence and the implications of assuming a limit exists.

Discussion Status

The conversation is ongoing, with various approaches being proposed. Some participants have suggested methods for proving convergence, while others are clarifying the assumptions and definitions involved in the problem. There is no explicit consensus yet, but several productive lines of reasoning have emerged.

Contextual Notes

Participants are navigating the complexities of the recursive definition and the implications of assuming convergence. There is mention of the need to show that the sequence is non-decreasing and bounded above, as well as the challenge of proving the existence of the limit.

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Homework Statement


Let [tex]s_{1} = 1[/tex] and [tex]s_{n+1} = \sqrt{s_{n} + 1}[/tex] Assume this converges to [tex]\frac{1}{2}(1+\sqrt{5})[/tex] and prove it.

Homework Equations


The Attempt at a Solution



I'm not really sure where to begin. I tried to set it up with [tex]|s_{n} - s| < \epsilon[/tex] but I'm not sure how to handle the recursiveness of the sequence.

Any tips would be appreciated.
 
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OK new attempt. Using algebra I found:
[tex]|\frac{2+s_{n}(1-\sqrt{5})}{2s_{n}}|< |\frac{2+M(1-\sqrt{5})}{2m}|<\epsilon[/tex]

where m = inf([tex]s_{n}[/tex]) and M = sup([tex]s_{n}[/tex])

Am I doing it right?
 
I'm not sure how you found that, since you didn't show it, but one part of the question is probably easier than you are making it out to be. If s_n has a limit L, then s_n->L and s_n+1->L. So L (if it exists) must satisfy L=sqrt(L+1). Can you solve that for L? What are the possibilities? Showing the limit actually exists is a little harder. Any ideas on how to go about it?
 
I'm not sure what you mean by "Assume this converges to [tex]\frac{1}{2}(1+ \sqrt{5})[/tex] and prove it".

Here's the proof reading that literally:
"If {sn} converges to [tex]\frac{1}{2}(1+ \sqrt{5})[/tex], then it converges to [tex]\frac{1}{2}(1+ \sqrt{5})[/tex]"!

If you mean "assume that it converges" and then prove that the limit is [tex]\frac{1}{2}(1+ \sqrt{5})[/tex], take the limit on both sides of [tex]s_{n+1}= \sqrt{s_n+ 1}[/tex]
Since the "s" terms on both sides give the same limit, if we call that limit L, we get [tex]L= \sqrt{L+ 1}[/itex] as Dick says.[/tex]
 
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If you need to show the sequence converges, you can do it by showing

1. The sequence is non-decreasing: [tex]s_1 \le s_2 \le \cdots \le s_n \cdots[/tex] (it's actually strictly increasing, but that's irrelevant)
2. The sequence is bounded above - there is a number [tex]W[/tex] such that [tex]s_n \le W[/tex] for all values of [tex]n[/tex].

Once you know it converges, you can proceed as the others have stated.

If you are free to ASSUME it converges then you can skip my two steps and do as they correctly suggest.
 
Blah, I did a terrible job of wording the problem. I think I may have gotten it, time will tell. Anyway, thank you for the help regardless! Next time I'll try better to word my questions. :)
 

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