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Thread: Sequence problem help please!

  1. #1
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    Dec 2007
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    Thumbs down Sequence problem help please!

    The given sequence is:
    {$\displaystyle a_n$}$\displaystyle = \sqrt{2+\sqrt{2+...+\sqrt{2}}} $, where n is the number of square roots, or in other words, $\displaystyle \{a_n\} = \sqrt{2+\{a_{n-1}\}} $
    In the first part, you solve to determine that {$\displaystyle a_{n+1}$} > {$\displaystyle a_n$}
    In the next part of the problem, it says to show by inducion that {$\displaystyle a_n$} is bounded above by 2 and then to find the $\displaystyle lim_{n\to\infty}${$\displaystyle a_n$}. But then I looked for sample problems in my workbook and found this exact question but it had said that it should be bounded by 3, not 2.

    Although it explicitly states that it should be 3, I can only get 2, I was wondering if I made an error or was it maybe a misprint in my book?

    {$\displaystyle a_n$} < 2 for all n = 1, 2, 3...
    Assume that it is true for n = k, that is, {$\displaystyle a_k$} < 2. Then,
    $\displaystyle \{a_{k+1}\} < 2 $
    $\displaystyle \sqrt{2+\{a_k\}}< 2 $
    However, $\displaystyle \{a_k\} < 2 $
    so $\displaystyle \sqrt{2+\{a_k\}} < \sqrt{2 + 2} < 2 $
    Therefore $\displaystyle \{a_k\}$ is always less than two.

    Then since it has been proved that this is a bounded monotonic sequence, it will converge to some value $\displaystyle \lim_{n\to\infty}\{a_n\} = L $
    This is how I solved:
    $\displaystyle \lim_{n\to\infty}\{a_{n+1}\} = L $

    $\displaystyle \lim_{n\to\infty}\sqrt{2+\{a_k\}} = L $

    $\displaystyle \frac{1}{2} \lim_{n\to\infty} (2+\{a_k\}) = L$

    $\displaystyle \frac{1}{2}(2 + L) = L $
    and then L = 2, so the sequence converges to 2.
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  2. #2
    Member
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    Mumbai
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    $\displaystyle Let y = sqrt(y+2)$ as n tends to infinity

    $\displaystyle y^2 = y+2$
    $\displaystyle y^2-y-2=0$
    solving for y,
    we get y=2 or y=-1,(which is impossible.)

    So the limit of the sequence should be 2.
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