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Math Help - Determine the convergence of a recursive sequence

  1. #1
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    Determine the convergence of a recursive sequence

    Let x_1 = 3 and x_{n+1} = (4-x_n)^{-1}

    Determine the convergence of the sequence.
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  2. #2
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    Can you show that this sequence is decreasing and bounded?
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  3. #3
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    Thanks Plato.

    We have 0<x_2 = 1 = \frac{1}{4-3}<3=x_1<4. Suppose 0<x_n<\cdots < x_1<4
    then x_{n+1} = 1/(4-x_n) > 0 and x_{n+1}-x_n = \frac{1}{4-x_n} -\frac{1}{4-x_{n-1}}=\frac{x_n-x_{n-1}}{(4-x_n)(4-x_{n-1})}<0
    Thus 0<x_{n+1}<x_n<\cdots < x_1<4 and so \{x_n\} is bounded and decreasing.
    Therefore it's convergent and L=\lim_{n \to \infty} x_n = \frac{1}{4-L}. Now L<x_2 =1 implies that L = 2-\sqrt{3}
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