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

  1. Oct 18, 2008 #1
    Find the limit as n tends to infinity;

    sqrt(2), sqrt(2+sqrt(2)), sqrt(2+sqrt(2+sqrt(2))), etc... n times
  2. jcsd
  3. Oct 18, 2008 #2


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    Re: limits

    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].
  4. Oct 19, 2008 #3
    Re: limits

    Thank you very much.
    All is clear.
  5. Oct 20, 2008 #4
    Re: limits

    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)
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