How would I find what a recursive sequence converges to a certain point?

In summary, the given sequence {a(n)}, refined recursively by a(1)=1 and a(n+1)=√(1+a(n)), is proven to converge through induction and its limit is found to be either (1+√5)/2 or (1-√5)/2. The latter is rejected because the sequence is increasing and the first solution is known as the golden ratio.
  • #1
Quantumpencil
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0

Homework Statement



Given the sequence {a(n)}, which is refined recursively by

a(1)=1, a(n+1)=√(1+a(n))

Prove that the sequence converges and find it's limit.

Homework Equations


The Attempt at a Solution



I proved that it converges with two pretty easy induction cases.

I've also found what it converges to, the golden ratio w/ a calculator. How would I go about this not using a calculator? Usually one would just consider the infinite case but here I'm not sure how to do that.
 
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  • #2
Hi Quantumpencil! :smile:

(have a square-root: √ :wink:)
Quantumpencil said:
a(1)=1, a(n+1)=sqrt(1+a(n))…
How would I go about this not using a calculator?

Just put n = n+1: a(n)=√(1+a(n)). :wink:
 
  • #3
Right, I got that; but don't you then end up with an infinite mess that's hard to evaluate?

a(n)=[tex]\sqrt{1+(a(n)}[/tex]; still forces you to substitute in a lot of a(n)'s; Is there no way to just take the limit and get phi?
 
  • #4
Quantumpencil said:
Right, I got that; but don't you then end up with an infinite mess that's hard to evaluate?

a(n)=[tex]\sqrt{1+(a(n)}[/tex]; still forces you to substitute in a lot of a(n)'s; Is there no way to just take the limit and get phi?

uhh? :confused: a(n) is fixed

if you prefer, just use a …

a = √(1 + a). :smile:
(and what happened to that √ that i gave you? :wink:)​
 
  • #5
What I'm saying is since each of the subsequent terms depends on the one before it... that sort of seems like magic to me. when you go to infinity there negligible difference between the n+1th and nth term in the sequence, but still...

Even with a(n)=[tex]\sqrt{1+a(n)}[/tex], what would I take a(n) to obtain the Golden ratio?

Sorry if I'm not making sense; I'm confused as to how making this equivalence helps us find the limit; we still don't have an (n) to take the limit with respect to...

Oh, and I noticed the sigma and clicked on it. Pretty cool stuff ;p
 
  • #6
Assume that the series converges.
Let [tex]lim_{n\rightarrow\infty}a(n)=l[/tex].
[tex]l = \sqrt{1+l}[/tex]
[tex]l^{2}-l-1=0[/tex]
[tex]l=(1+\sqrt{5})/2[/tex] or [tex](1-\sqrt{5})/2[/tex]
 
  • #7
I have to disagree with tiny-tim here, at least with the way he said it. Saying "put n= n+1" is exactly the same as saying 0= 1! What he meant, I'm sure, is that the sequence {a(n)} necessarily has the same limit as the sequence {a(n+1)}, the same sequence, "shifted" by 1. That leads to what evail is saying:
[tex]\lim_{n\rightarrow\infty} a(n)= \sqrt{1+ \lim_{x\rightarrow\infty} a(n+1)}[/tex]
so if the limit is "[itex]l[/itex]", [itex]l= \sqrt{1+ l}[/itex] which has the solutions evail shows.

Of course, the limit is one of [itex](1+ \sqrt{5})/2[/itex] or [itex](1- \sqrt{5})/2[/itex]. Do you see how to determine which one? a(1)= 1>0 and it is easy to see that this sequence is increasing. The limit can't be negative!
 
  • #8
HallsofIvy said:
I have to disagree with tiny-tim here, at least with the way he said it. Saying "put n= n+1" is exactly the same as saying 0= 1! What he meant, I'm sure, is that the sequence {a(n)} necessarily has the same limit as the sequence {a(n+1)}, the same sequence, "shifted" by 1. That leads to what evail is saying:
[tex]\lim_{n\rightarrow\infty} a(n)= \sqrt{1+ \lim_{x\rightarrow\infty} a(n+1)}[/tex]
so if the limit is "[itex]l[/itex]", [itex]l= \sqrt{1+ l}[/itex] which has the solutions evail shows.

Of course, the limit is one of [itex](1+ \sqrt{5})/2[/itex] or [itex](1- \sqrt{5})/2[/itex]. Do you see how to determine which one? a(1)= 1>0 and it is easy to see that this sequence is increasing. The limit can't be negative!

Thanks HallsofIvy for completing my solution! i forgot that a(n+1)>a(n) for all n>1. hahaha.. so i had to reject the negative part :)
 

1. How do I determine the convergence of a recursive sequence?

To determine the convergence of a recursive sequence, you can use the concept of a limit. Take the limit of the recursive sequence as n approaches infinity, and if the limit exists and is a finite number, then the sequence is convergent.

2. Can I use a formula to find the convergence of a recursive sequence?

Yes, there are certain formulas that can be used to find the convergence of a recursive sequence. One common formula is the geometric series formula, which is commonly used for geometric sequences. However, not all recursive sequences have a formula that can be used to find their convergence.

3. How does the initial value affect the convergence of a recursive sequence?

The initial value, also known as the starting term, can greatly affect the convergence of a recursive sequence. If the initial value is close to the convergence point, the sequence will converge faster. However, if the initial value is far from the convergence point, the sequence may take longer to converge or may not converge at all.

4. Can I use a graph to visualize the convergence of a recursive sequence?

Yes, a graph can be a helpful tool in visualizing the convergence of a recursive sequence. Plotting the terms of the sequence on a graph and connecting them with a line can show the trend of the sequence and whether it is approaching a certain point or oscillating around it.

5. Are there any techniques for finding the convergence point of a recursive sequence?

Yes, there are several techniques that can be used to find the convergence point of a recursive sequence. Some common techniques include using a formula, using a graph, or solving the recursive formula for a specific term and taking the limit as n approaches infinity. However, the most reliable way to find the convergence point is to use the concept of a limit.

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