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Math Help - Limit (1)

  1. #1
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    Limit (1)

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


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

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

    x=ln(1+x)

    which is true only for x=0.

    But \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 x_n \to 0 while n \to \infty.
    By intuition nx_n tends to 0 as 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 \ln (x) for x=1 is 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; June 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

    ln(r)<r for all real numbers

    then the sequence

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

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

    lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=lop= 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

    lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=lop= 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 \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 \lim_{n\to \infty}\frac{1}{x_{n}}=\infty

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

    x_{n+1}<x_{n}<x_{n-1}<.......<x_1

    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 \lim_{n\rightarrow\infty} \frac{1}{x_{n}} = \infty
    (i'm not sure,i'm just trying to help)
    Last edited by Raoh; June 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

    x_{n+1}<x_{n}<x_{n-1}<.......<x_1

    x_{n+1}=ln(x_{n})<x_{n}

    I am not sure about how I derive it ..
    Caution: 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: x_{n+1}=\ln (1+x_n).
    It still the same

    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 \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 \{y_n \} if \lim_{n\to\infty} y_n =y, then \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
    x_{n}\rightarrow 0  while n \rightarrow 0
    (thanks to "arbolis" )
    then
    \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 \{y_n \} if \lim_{n\to\infty} y_n =y, then \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: <br />
\mathop {\lim }\limits_{n \to  + \infty } x_n  = 0<br />

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


    Part 2: <br />
\mathop {\lim }\limits_{n \to  + \infty } \tfrac{{x_{n + 1} }}<br />
{{x_n }} = 1<br />

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


    Part 3: <br />
\left( {\tfrac{1}<br />
{{x_{n + 1} }} - \tfrac{1}<br />
{{x_n }}} \right) \to \tfrac{1}<br />
{2}<br /> <br />

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


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

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

    (*) A discrete version of L'Hôpital
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