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Series Problem

  1. Jul 31, 2007 #1
    1. The problem statement, all variables and given/known data

    a1 = sqrt(3), a2 = sqrt( 3 + sqrt( 3) ), a3 = sqrt( 3 + sqrt( 3 + sqrt( 3) ) )

    2. Relevant equations

    Notice that each term is inside the sqrt of the previous term. I have no idea how to lay something like this out. Any help would be greatly appreciated.

    3. The attempt at a solution
     
  2. jcsd
  3. Jul 31, 2007 #2
    What are you trying to do? Find an expression for the nth term? Find the sum of the series? What?
     
  4. Jul 31, 2007 #3
    sorry,

    it says find a recursive formula for a n+1 in terms of an. and find its limit
     
  5. Jul 31, 2007 #4
    [tex]a_{n+1}=\sqrt{3+a_n}[/tex]

    We have [tex]a_1<a_2[/tex]
    Suppose [tex]a_{n-1}<a_n[/tex].
    Then [tex]a_n-a_{n+1}=\sqrt{3+a_{n-1}}-\sqrt{3+a_n}<0\Rightarrow a_n<a_{n+1}[/tex], so [tex](a_n)_{n\geq 1}[/tex] is crescent.
    We'll prove that [tex]\displaystyle a_n<\frac{1+\sqrt{13}}{2}[/tex]
    [tex]a_1<\sqrt{1+\sqrt{13}}{2}[/tex].
    Suppose that [tex]a_n<\frac{1+\sqrt{13}}{2}[/tex].
    Then [tex]a_{n+1}=\sqrt{3+a_n}<\sqrt{3+\frac{1+\sqrt{13}}{2}}=\frac{1+\sqrt{13}}{2}[/tex].
    So the sequence is convergent. Let [tex]l=\lim_{n\to\infty}a_n[/tex].
    Then [tex]l=\sqrt{3+l}\Rightarrow l=\frac{1+\sqrt{13}}{2}[/tex]
     
  6. Jul 31, 2007 #5
    Thanks but I'm more confused now that I thought I was before. Where did your sqrt(13) come from?

    thanks,

    glenn
     
  7. Jul 31, 2007 #6
    Nevermind, I understand now. Thanks for the help...
     
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