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Math Help - Recursion sequence, and its limit

  1. #1
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    Recursion sequence, and its limit

    Let a,c>0, a sequence { a_n} is defined:

    a_1=c

    a_{n+1}=\sqrt{a+a_n}

    (1) For which values of c the sequence gets lower, higher or does not change?
    (2) Is there a limit in each case? Find it.

    Okay, in the first case it's easy, because then a_n=c for every n, and of course the limit is c.
    Only I found some weird expressions for c depending on a:
    I found that a_{n+1}>a_n when: c^2-c<a, which I don't know how to turn it into a 'nicer' expression of c.

    That's before trying to find the limit of them...

    Thank you very much!
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  2. #2
    MHF Contributor chisigma's Avatar
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    If You suppose that the sequence a_{n} tends to a finite limit, the limit itself is obtained imposing that...

    \lim_{n \rightarrow \infty} a_{n+1} - a_{n}=0 (1)

    ... i.e. the solution of the equation...

    x=\sqrt{a+x} (2)

    ... so that is...

    a_{\infty} = \frac{1 + \sqrt{1+4a}}{2} (3)

    It is important to consider that the limit (3) is independent from a_{1}=c if c=0 so that...

    a) if a_{1}<a_{\infty} then the sequence is monotonically increasing...

    b) if a_{1}=a_{\infty} then the sequence is constant...

    c) if a_{1}>a_{\infty} then the sequence is monotonically decreasing ...

    Kind regards

    \chi \sigma
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