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Thread: Recursion sequence, and its limit

  1. #1
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    Recursion sequence, and its limit

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

    $\displaystyle a_1=c$

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

    (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 $\displaystyle a_n=c$ for every n, and of course the limit is c.
    Only I found some weird expressions for c depending on a:
    I found that $\displaystyle a_{n+1}>a_n$ when: $\displaystyle c^2-c<a$, 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 $\displaystyle a_{n}$ tends to a finite limit, the limit itself is obtained imposing that...

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

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

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

    ... so that is...

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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