Prove that the sequence converges and find its limit

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Discussion Overview

The discussion revolves around proving the convergence of a sequence defined by a recurrence relation, specifically starting with a_1 = 2^(1/2) and a_n = (2 + a_n-1)^(1/2). Participants explore methods to establish convergence and identify the limit of the sequence.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest guessing the limit based on the recurrence relation and then proving convergence by showing the sequence is increasing and bounded.
  • One participant notes the sequence appears to be increasing but expresses uncertainty about how to demonstrate it is bounded, questioning the implications if a_n reaches a certain value.
  • Another participant proposes writing out the first few terms to help identify an upper bound.
  • A later reply introduces a specific value for the limit using a cosine relationship, which is met with both agreement and skepticism regarding the increasing nature of the terms involved.
  • Induction is presented as a method to show that the sequence remains less than 2, supporting the claim of convergence.

Areas of Agreement / Disagreement

Participants generally agree that the sequence is increasing, but there is disagreement about how to establish its boundedness and the specific limit. Multiple competing views on the limit and the methods to prove convergence are present.

Contextual Notes

There are unresolved questions regarding the upper bound of the sequence and the implications of certain values that a_n might take. The discussion includes various approaches and assumptions that have not been fully validated.

mansi
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given a recurrence relation, a_1 =2^(1/2) and a_n = (2 +a_n-1)^1/2 ...prove that the sequence converges and find its limit..
are we supposed to begin by guessing the limit and the bounds ??
 
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Yes, try to guess the limit. You can make an educated guess though, if you assume that your sequence converges you should be able to find what it's limit must be (hint:use the recurrance relation).

Once you've got this 'guess' limit in hand, you should be able to prove convergence (hint:show it's an increasing & bounded sequence).
 
By mere observation, it's quite clear that a_1 < a_2 <...<a_n...
so, it's an increasing sequence...
but i can't think of how we can show it's bounded...i mean,how do we use the recurrence relation?..and i guess, once we find the upper bound it would be easy to spot the limit of the sequence...
 
Maybe you could write out the first few terms, that might hint towards an upper bound.
 
mansi said:
By mere observation, it's quite clear that a_1 < a_2 <...<a_n...
so, it's an increasing sequence...

You don't think you need any conditions on the terms to deduce this? What if, at some point, [tex]a_n=3[/tex]. Then what's [tex]a_{n+1}[/tex]? Can you guarantee this won't hapen?

mansi said:
..and i guess, once we find the upper bound it would be easy to spot the limit of the sequence...

If you happened to find the least upper bound, then yes.
 
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Robert,it can't be.The #-s on the right (the square roots) are increasing,while the sinus-es are decreasing (their arguments tends to 0)...

Daniel.

EDIT:You edited,and put "cos".Now it makes sense.Both are increasing.This "cosine" trick is really elegant... :approve:
 
Last edited:
dextercioby: This "cosine" trick is really elegant.:

Wow! Thanks for the complement. I at first mistakenly put in the sin, which has this form:

2sin x = [tex]\sqrt{2-\sqrt{2+\sqrt{2+++}}}[/tex]

So that we get sin^2 + cos^2 = 1.
 
Last edited:
Induction is a pretty simple way to show that the sequence converges:
A1 = 2^(1/2) < 2
Assume An < 2
A(n+1) = (An + 2)^(1/2) < (2 + 2)^(1/2) = 2
By the principle of mathematical induction, An < 2 for all n.
 

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