Find sqrt( 2 + sqrt( 2 + sqrt( 2 + )))

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Homework Help Overview

The discussion revolves around finding the limit of a recursive sequence defined by nested square roots, specifically the expression sqrt(2 + sqrt(2 + sqrt(2 + ...))). The subject area includes concepts from calculus and sequences.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the convergence of the sequence defined by a recursive formula and explore the implications of the limit. There is an attempt to establish the relationship A = sqrt(2 + A) based on the limit of the sequence.

Discussion Status

The discussion is active, with some participants providing guidance on proving convergence and questioning the reasoning behind the limit relationship. There is a mix of understanding and confusion regarding the definitions and implications of the limit.

Contextual Notes

Participants are considering the need to prove that the sequence is increasing and bounded, which are key aspects of establishing convergence. There is also a noted confusion about the relationship between the limit of the sequence and its recursive definition.

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Find the limit as n tends to infinity;

sqrt(2), sqrt(2+sqrt(2)), sqrt(2+sqrt(2+sqrt(2))), etc... n times
 
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If an is the nth value, then [itex]a_{n+1}= \sqrt{2+ a_n}[/itex]

IF the sequence {an} converges (you will need to prove that, perhaps by proving that it is an increasing sequence and has an upper bound), call the limit A.
Then we must have [itex]\lim_{n\rightarrow \infty}a_{n+1}= \sqrt{2+ \lim_{n\rightarrow \infty}a_n}[/itex] so [itex]A= \sqrt{2+ A}[/itex].
 


Thank you very much.
All is clear.
 


Sorry i have a small problem

How come A=sqrt(2+A) ?
A=limit of an not a(n+1)
so Limit a(n+1)=sqrt(2+A)
No?
 

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