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Recursive sequence question

  1. Sep 3, 2014 #1
    I want to prove that
    if the sequence [itex]a_n [/itex] satisfy that
    [itex]a_{n+1}=a_n\left(1-c\frac{a_n}{1+a_n}\right)[/itex]
    then [itex]a_n*c*n\rightarrow 1[/itex] for all positive [itex]c[/itex].

    Like when [itex]c=1[/itex], then [itex]a_n*n\rightarrow 1[/itex],
    but if [itex]c\neq 1[/itex], it's difficult to prove.
     
  2. jcsd
  3. Sep 3, 2014 #2

    mathman

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    What makes you believe it is true? Your question implies [itex]a_n*n ->\frac{1}{c}[/itex]. Doesn't look right, especially for large c.
     
    Last edited: Sep 3, 2014
  4. Sep 3, 2014 #3
    Thank you for you reply.
    I think it's correct.
    First, I can have [itex]a_n*n\rightarrow 1[/itex] when [itex]c=1[/itex], from
    [itex]a_{n+1}=\frac{a_n}{1+a_n}=\frac{a_{n-1}}{1+2a_{n-1}}=\cdots=\frac{a_1}{1+(n+1)a_1}[/itex]

    Then, let [itex]ca_n=b_n [/itex] if [itex]c\neq 1[/itex], [itex]c[/itex] is some constant. we can have [itex]b_{n+1}=b_n\left(1-\frac{b_n}{1+b_n/c}\right)[/itex].

    Actually, it's easy to prove [itex]a_n[/itex] and [itex]b_n[/itex] will go to zero,
    so, [itex]\frac{b_n}{1+b_n/c}\sim\frac{b_n}{1+b_n}[/itex], then[itex] b_n*n\rightarrow 1[/itex].

    I have made a plot, it's correct no matter [itex]c[/itex] is larger or less than 1.
    But I think my proof is not strict.
    Thanks for your concern.
     
  5. Sep 3, 2014 #4
    Oh, I forget there is a condition that
    the sequence should satisfy that[itex]1-c\frac{a_1}{1+a_1}[/itex]>0,
    so that all the elements in this sequence should be positive, and c cannot be too large.
    I have made some plots like [itex]c=0.5, c=2[/itex], the conclusion is correct.
    Thanks
     
  6. Sep 4, 2014 #5

    mathman

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    Write out your complete proof.
     
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