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Math Help - Convergence

  1. #1
    Member thaopanda's Avatar
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    Convergence

    Consider the sequence of real number { a_n} n \in N defined recursively by a_1 = 0; a_n+1 = \sqrt{6+a_n}   \forall n \geq 1

    Show that the sequence is convergent and find its limit.
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  2. #2
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    Quote Originally Posted by thaopanda View Post
    Consider the sequence of real number { a_n} n \in N defined recursively by a_1 = 0; a_n+1 = \sqrt{6+a_n} \forall n \geq 1

    Show that the sequence is convergent and find its limit.

    1.- prove inductively that the sequence is bounded above , by 3 say.

    2.- prove inductively that a_n <= a_(n+1) (monot. ascending)

    3.- thus the limit exists by a theorem. Now use a_(n+1) = sqrt(6 + a_n) and arithmetic of limits to find the limit.

    Tonio
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  3. #3
    Member thaopanda's Avatar
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    wow, I was actually really close to that. I started to use induction, but then I thought I had to prove it was cauchy or something since that's all my professor forces into our head.... thanks!
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