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Find the limit of the sequence

  1. Oct 17, 2004 #1

    quasar987

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    ... [itex]\{x_n\}[/itex] defined by [itex]x_1=1[/itex] and [itex]x_{n+1}=3+\sqrt{x_n}[/itex] for [itex]n\geq 2[/itex].

    Here's what I did. We know for sure that a sequence as a limit if either it it increasing and has a superior bound or if it is decreasing and has an inferior bound. So let's suppose it satisfy either one of these condition and let's see what are the posible candidates for the limit. We supose that

    [tex]\lim_{n\rightarrow \infty} x_n=x[/tex]

    Now the limit of [itex]x_{n+1}[/itex] must [itex]x[/itex] too, because [itex]\{x_{n+1}\}[/itex] is a sub-sequence of [itex]\{x_n\}[/itex]. But we can find another expression for the limit of [itex]x_{n+1}[/itex], that is,

    [tex]\lim_{n\rightarrow \infty} x_{n+1}=\lim_{n\rightarrow \infty} 3+\sqrt{x_n}= 3 + \sqrt{\lim_{n\rightarrow \infty} {x_n}}= 3 + \sqrt{x}[/tex]

    So for the limit to be unique, as suggested by our hypothesis, we must have

    [tex]x=3 + \sqrt{x}[/tex]

    Now how do you find the roots of this equation? I tried to set [itex]\sqrt{x}=y[/itex], so that the expression in x becomes

    [tex]y^2=3 + y \Leftrightarrow y^2-y-3=0[/tex]

    but this yields solutions

    [tex]x=\left(\frac{1\pm\sqrt{13}}{2} \right)^2[/tex]

    The manual gives

    [tex]\frac{7+\sqrt{13}}{2}[/tex]

    as the answer to the problem. It's close but at the same time not. How would you go about this problem?
     
    Last edited: Oct 18, 2004
  2. jcsd
  3. Oct 17, 2004 #2

    Hurkyl

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    You sure it's not?
     
  4. Oct 17, 2004 #3

    quasar987

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    The posters should be allowed to delete they threads... shame on me. :shy:

    But thanks again Hurkyl.
     
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