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a sequence {an} is given by a1 = (3)^1/2 , an+1 = (3+an)^1/2. Show that {an} is monotonic and bounded, and calculate the limit if it exists

  • Math -

    A(n) = √(3 + √(3 + √(3 + ...))..) to infinity
    let x = √(3 + √(3 + √(3 + ...))..)
    square both sides
    x^2 = 3 + √(3 + √(3 + √(3 + ...))..)
    or
    x^2 = 3 + x
    x^2 - x - 3 = 0
    by the formula
    x = (1+√13)/2
    = appr. 2.302775638... which I got by repeatedly doing the following on my calculator about 20 times

    1. √3
    2. =
    3. +3
    4. =
    5. √
    goto 2.

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