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Math Help - recursive sequence

  1. #1
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    recursive sequence

    hi,
    i need to find the limit of the following sequence
    lim (an)1/n
    n
    where an the sequence defined recursively as follows
    a1=1 and an+1=n(1-lnan) where ln is the natural logatithm
    I can prove (using induction) that the sequence is increasing and not upper bounded but I don't know how to prove that. I'm sorry but I don't know how to write the sequence in the correct form, so I attached a doc file
    lim (an)1/n=1 n

    Attached Files Attached Files
    Last edited by nolimits; April 27th 2009 at 09:09 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting...

    c_{n}= (a_{n})^{\frac{1}{n}}

    ... where the sequence a_{n} is defined as...

    a_{1}=1, a_{n+1}=n\cdot (1-\ln a_{n})

    ... after easy steps you obtain...

     \ln c_{n}= \frac{\ln (n-1)  + \ln (1-\ln a_{n-1})}{n}< \frac{\ln (n-1)}{n} (1)

    From (1) it derives that...

    \lim_{n \rightarrow \infty} \ln c_{n}=0

    ... so that...

    \lim_{n \rightarrow \infty} c_{n}=1

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 26th 2009 at 11:55 PM. Reason: minor corrections
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  3. #3
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    thanks  \chi  \sigma for your answer but i've seen I made a mistake while writing the sequence.
    the correct sequence is
    a_{1}=1   \ a_{n+1} = n \cdot(1+\ln a_{n})
    I'm sorry
    Last edited by nolimits; April 27th 2009 at 09:08 AM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    In that case we have to find...

    \lim_{n \rightarrow \infty} c_{n} (1)

    ... where...

    c_{n}= (a_{n})^{\frac{1}{n}}= e^{\frac{\ln a_{n}}{n}} (2)

    ... and the sequence a_{n} is defined as...

    a_{1}=1, a_{n+1}= n\cdot (1+\ln a_{n}) (3)

    In this case may help a preliminary test in order to undestand the behavior of the term \ln a_{n} in (2). Value of \ln a_{n} and \ln n computed till to n=1000 are plotted in figure...



    Setting...

    \delta_{n}= \ln a_{n} - \ln n (4)

    ... il seems that the sequence \delta_{n} is upper bounded, that means that...

    \lim_{n \rightarrow \infty} \delta_{n}= \delta (5)

    ... where 2<\delta<3. In this case is...

    \lim_{n\rightarrow \infty} \frac{\ln a_{n}}{\ln n} = 1 (6)

    ... so that...

    \lim_{n \rightarrow \infty} \frac{\ln a_{n}}{n}= \lim_{n \rightarrow \infty} \frac{\ln n}{n}= 0 \rightarrow \lim_{n \rightarrow \infty} c_{n}=1 (7)

    All that however is not yet demonstrated ...

    Here there are some value of \delta_{n} with n increasing...

    \delta_{1000}= 2.324429\dots

    \delta_{10000}= 2.5459131\dots

    \delta_{50000}=2.6736756\dots

    \delta_{100000}=2.72369079\dots

    \delta_{200000}=2.771159376\dots

    Kind regards

    \chi \sigma
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