Finding the Limit of a Series: [a1]=√12, [an+1]=√(12+an)

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Homework Help Overview

The discussion revolves around the convergence of a series defined by the recursive relation [a1]=√12 and [an+1]=√(12+an). Participants are exploring the limit that the series approaches as n tends to infinity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to express the limit L in terms of the recursive definition and are questioning the nature of convergence. There are discussions about substituting L into the recursive formula and exploring the implications of convergence.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts on the convergence of the series and the limit it approaches. Some have made attempts to derive expressions for L, while others are seeking clarification on the limit itself.

Contextual Notes

There is a lack of consensus on the exact value of the limit, and participants are navigating through the implications of the recursive relationship without a definitive conclusion.

jaqueh
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Homework Statement


What does the series converge to?
[a1]=√12
[an+1]=√(12+an)


Homework Equations


Let L = the limit it approaches


The Attempt at a Solution


I don't know if i did this correctly but I made
L = √(12+√(12+√(12+...)))
then L2 = 12+√(12+√(12+...))
then L = 12 + 12∞-1...12
thus L = ∞√(12 + 12∞-1...12)
 

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If it converges then a_n approaches L as n->infinity. So does a_(n+1). Put that into a_n=sqrt(12+a_(n+1)).
 
Dick said:
If it converges then a_n approaches L as n->infinity. So does a_(n+1). Put that into a_n=sqrt(12+a_(n+1)).

ok i get that an+1 converges now because it is in the an series, but what is the limit that it approaches?
 
jaqueh said:
ok i get that an+1 converges now because it is in the an series, but what is the limit that it approaches?

It approaches the same limit as a_n, call it L. Solve for L!
 

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