Recursive sequence convergence

In summary, the author is trying to solve a problem with a sequence and is having difficulty. They have tried different methods and found a limit. If they are free to assume the sequence converges, they can skip the two steps and get to the solution.
  • #1
SticksandStones
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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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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
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  • #5
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.
 
  • #6
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. :)
 

FAQ: Recursive sequence convergence

1. What is a recursive sequence convergence?

A recursive sequence convergence is a sequence of numbers where the next term in the sequence is defined in terms of the previous terms. The sequence is said to converge if the terms get closer and closer to a specific value as the sequence progresses.

2. How do you determine if a recursive sequence converges?

To determine if a recursive sequence converges, you can use the method of induction or the squeeze theorem. The method of induction involves proving that the sequence is bounded and monotonic, while the squeeze theorem involves finding two other sequences that have the same limit as the original sequence.

3. What is the difference between a convergent and a divergent recursive sequence?

A convergent recursive sequence is one that has a limit, meaning that the terms in the sequence get closer and closer to a specific value as the sequence progresses. A divergent recursive sequence, on the other hand, does not have a limit and the terms in the sequence can either increase or decrease without bound.

4. Can a recursive sequence converge to more than one value?

No, a recursive sequence can only converge to a single value. This is because a convergent sequence is defined as one where the terms get closer and closer to a specific value as the sequence progresses. If the sequence were to converge to more than one value, the terms would not be getting closer and closer to a single value.

5. Are there any real-life applications of recursive sequence convergence?

Yes, recursive sequence convergence has various real-life applications, such as in finance, physics, and computer science. In finance, it can be used to model compound interest or stock market fluctuations. In physics, it can be used to model the motion of objects under the force of gravity. In computer science, it can be used to optimize algorithms and solve problems in data structures.

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