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Thread: Limit (1)

  1. #1
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    Limit (1)

    Suppose $\displaystyle x_1 > 0$ is given and let $\displaystyle x_{n+1}=\ln(1 + x_n), \ \ n \geq 1.$ Evaluate $\displaystyle \lim_{n\to\infty} nx_n.$


    Spoiler:
    The answer does not depend on the value of $\displaystyle x_1.$
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  2. #2
    Senior Member pankaj's Avatar
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    Let $\displaystyle \lim_{n\to \infty}x_{n}=x$

    $\displaystyle \lim_{n\to \infty}x_{n+1}=\lim_{n\to \infty}(ln(1+x_{n}))$

    $\displaystyle x=ln(1+x)$

    which is true only for $\displaystyle x=0$.

    But $\displaystyle \lim_{n\to \infty}nx_{n}$ is something like
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  3. #3
    MHF Contributor arbolis's Avatar
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    If I'm not wrong the indetermination arises because $\displaystyle x_n \to 0$ while $\displaystyle n \to \infty$.
    By intuition $\displaystyle nx_n$ tends to $\displaystyle 0$ as $\displaystyle n$ tends to positive infinite. However I'm not sure at all, and I guess it could tend to a constant. This is a tough problem!

    Or is it 1 because the slope of $\displaystyle \ln (x)$ for $\displaystyle x=1$ is $\displaystyle 1$...?

    Edit : Now my question is... can we prove it with Peano's axioms? Or we have to suppose another set of axioms?
    I'm sad I don't have enough time to dedicate myself fully to the problem.
    Last edited by arbolis; Jun 9th 2009 at 07:36 AM.
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  4. #4
    Senior Member pankaj's Avatar
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    I also think that answer should be zero,but a rigorous solution must be there.
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  5. #5
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    it's too soon to give away any part of solution. so i'll just give you a probably useless hint!

    Spoiler:
    the answer is:


    Spoiler:
    2
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  6. #6
    MHF Contributor Amer's Avatar
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    since

    $\displaystyle ln(r)<r $ for all real numbers

    then the sequence

    $\displaystyle x_{n+1}=ln(x_n) $ decreasing so the limit of it zero

    $\displaystyle lim_{n\rightarrow \infty } nx_n =lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=\frac{\infty}{\infty}$ use lopitals

    $\displaystyle lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=lop=$$\displaystyle lim_{n\rightarrow \infty } \frac{1}{\frac{(x'_{n-1})}{(x_{n-1})(ln(x_{n-1})^2)}}=lim_{n\rightarrow \infty }\frac{(x_{n-1})(ln(x_{n-1})^2)}{(x'_{n-1})}=lop..lop...lop..lop=x_1$

    I think my solution wrong right
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  7. #7
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Amer View Post
    since
    [/tex] use lopitals

    $\displaystyle lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=lop=$$\displaystyle lim_{n\rightarrow \infty } \frac{1}{\frac{(x'_{n-1})}{(x_{n-1})(ln(x_{n-1})^2)}}=lim_{n\rightarrow \infty }\frac{(x_{n-1})(ln(x_{n-1})^2)}{(x'_{n-1})}=lop..lop...lop..lop=x_1$

    I think my solution wrong right
    How do you know that $\displaystyle \lim_{n\to \infty}\frac{1}{x_{n}}=\infty$

    Therefore using L'Hopital rule is not proper
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  8. #8
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by pankaj View Post
    How do you know that $\displaystyle \lim_{n\to \infty}\frac{1}{x_{n}}=\infty$

    Therefore using L'Hopital rule is not proper
    since it is decreasing sequence

    $\displaystyle x_{n+1}<x_{n}<x_{n-1}<.......<x_1$

    $\displaystyle x_{n+1}=ln(x_{n})<x_{n}$

    I am not sure about how I derive it ..
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  9. #9
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    Smile

    even if it's decreasing,we can't say that $\displaystyle \lim_{n\rightarrow\infty} \frac{1}{x_{n}} = \infty$
    (i'm not sure,i'm just trying to help)
    Last edited by Raoh; Jun 9th 2009 at 04:20 PM.
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  10. #10
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Amer View Post
    since it is decreasing sequence

    $\displaystyle x_{n+1}<x_{n}<x_{n-1}<.......<x_1$

    $\displaystyle x_{n+1}=ln(x_{n})<x_{n}$

    I am not sure about how I derive it ..
    Caution: $\displaystyle x_{n+1}=\ln (1+x_n)$.
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  11. #11
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by arbolis View Post
    Caution: $\displaystyle x_{n+1}=\ln (1+x_n)$.
    It still the same

    $\displaystyle x_n>ln(x_{n}+1)$



    the sequence is decreasing as I said try it
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  12. #12
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    here are two important pieces of the solution. see if you can put everything together and give a complete solution:

    1) find $\displaystyle \lim_{n\to\infty} \left(\frac{1}{x_{n+1}}-\frac{1}{x_n} \right).$

    2) it is a well-known result that for any sequence $\displaystyle \{y_n \}$ if $\displaystyle \lim_{n\to\infty} y_n =y,$ then $\displaystyle \lim_{n\to\infty} \frac{y_1+y_2+ \cdots + y_n}{n}=y.$
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  13. #13
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    Smile

    hi
    $\displaystyle x_{n}\rightarrow 0 $ while $\displaystyle n \rightarrow 0$
    (thanks to "arbolis" )
    then
    $\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1}{x_{n+1}}-\frac{1}{x_{n}} \right)= \frac{1}{2}$
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  14. #14
    Moo
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    Quote Originally Posted by NonCommAlg View Post

    2) it is a well-known result that for any sequence $\displaystyle \{y_n \}$ if $\displaystyle \lim_{n\to\infty} y_n =y,$ then $\displaystyle \lim_{n\to\infty} \frac{y_1+y_2+ \cdots + y_n}{n}=y.$
    ...called Cesaro's mean, for those who're interested
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  15. #15
    Super Member PaulRS's Avatar
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    Part 1: $\displaystyle
    \mathop {\lim }\limits_{n \to + \infty } x_n = 0
    $

    Spoiler:
    The sequence is decreasing and positive, hence it converges, let $\displaystyle L$ be its limit then: $\displaystyle
    \log \left( {1 + L} \right) = L
    $ and $\displaystyle L\geq{0}$ (since our sequence is positive), but this can happen only when $\displaystyle L=0$ ( See that $\displaystyle
    f\left( x \right) = x - \log \left( {1 + x} \right)
    $ is increasing in $\displaystyle
    \left( {0, + \infty } \right)
    $ )


    Part 2: $\displaystyle
    \mathop {\lim }\limits_{n \to + \infty } \tfrac{{x_{n + 1} }}
    {{x_n }} = 1
    $

    Spoiler:
    $\displaystyle
    \tfrac{{x_{n + 1} }}
    {{x_n }} = \tfrac{{\log \left( {1 + x_n } \right)}}
    {{x_n }} = \tfrac{{x_n + o\left( {x_n } \right)}}
    {{x_n }} = 1 + o\left( 1 \right)
    $ as $\displaystyle n\to +\infty$ because $\displaystyle x_n\to {0^+}$


    Part 3: $\displaystyle
    \left( {\tfrac{1}
    {{x_{n + 1} }} - \tfrac{1}
    {{x_n }}} \right) \to \tfrac{1}
    {2}

    $

    Spoiler:
    By Taylor (our sequence converges to 0): $\displaystyle
    \left( {\tfrac{1}
    {{x_{n + 1} }} - \tfrac{1}
    {{x_n }}} \right) = - \tfrac{{\log \left( {1 + x_n } \right) - x_n }}
    {{x_n x_{n + 1} }} = - \left( {\frac{{ - \tfrac{{x_n ^2 }}
    {2} + o\left( {x_n ^2 } \right)}}
    {{x_n x_{n + 1} }}} \right) = \tfrac{{x_n }}
    {{2x_{n + 1} }} + o\left( {\tfrac{{x_n }}
    {{x_{n + 1} }}} \right)
    $ as $\displaystyle n\to +\infty$ and we are done by Part 2


    Part 4:
    Computing $\displaystyle \lim_{n\to +\infty}{n\cdot x_n}$

    Spoiler:
    Since $\displaystyle
    \mathop {\lim }\limits_{n \to + \infty } \tfrac{{\left( {n + 1} \right) - n}}
    {{\tfrac{1}
    {{x_{n + 1} }} - \tfrac{1}
    {{x_n }}}} = 2
    $ by Part 3 , $\displaystyle x_n$ decreases and $\displaystyle
    \mathop {\lim }\limits_{n \to + \infty } \tfrac{1}
    {{x_n }} = + \infty
    $, by the Stolz-Cesaro Theorem (*) we have that $\displaystyle
    2 = \mathop {\lim }\limits_{n \to + \infty } \tfrac{n}
    {{\tfrac{1}
    {{x_n }}}} = \mathop {\lim }\limits_{n \to + \infty } n \cdot x_n
    $

    (*) A discrete version of L'Hôpital
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