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Math Help - Strictly contracting sequences

  1. #1
    Super Member Showcase_22's Avatar
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    Strictly contracting sequences

    if (a_n) is strictly contracting then \sum_{n=1}^\infty |a_{n+1}-a_n| converges.

    Is this statement true or false?
    I decided that this statement was true but I couldn't find a concrete way of justifying it.

    I tried using |a_{n+1}-a_n| \leq L|a_n-a_{n-1}| but i'm not sure how that really helps.
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  2. #2
    Moo
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    Hello !
    Quote Originally Posted by Showcase_22 View Post
    I decided that this statement was true but I couldn't find a concrete way of justifying it.

    I tried using |a_{n+1}-a_n| \leq L|a_n-a_{n-1}| but i'm not sure how that really helps.
    And remember that {\color{red}0<L<1}

    Here is the general idea :
    \sum_{n=1}^\infty |a_{n+1}-a_n|=\lim_{N \to \infty} S_N, where S_N=\sum_{n=1}^N |a_{n+1}-a_n|

    It can easily be shown (by induction) that |a_{n+1}-a_n| \leq L^{n-1} |a_2-a_1|

    Hence \sum_{n=1}^N |a_{n+1}-a_n| \leq \sum_{n=1}^N L^{n-1} |a_2-a_1|=|a_2-a_1| \sum_{n=1}^N L^{n-1}=|a_2-a_1| \sum_{n=0}^{N-1} L^n


    Therefore \sum_{n=1}^\infty |a_{n+1}-a_n|=\lim_{N \to \infty} |a_2-a_1| \sum_{n=0}^{N-1} L^n=|a_2-a_1| \lim_{N \to \infty} \sum_{n=0}^{N-1} L^n

    Is it a converging series ?



    Note : I introduced N, because since we don't know if it converges or not, we're normally not allowed to write the infinity in the sum
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  3. #3
    Super Member Showcase_22's Avatar
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    |a_2-a_1|\lim_{N \rightarrow \infty}\sum_{n=0}^{N-1}L^n

    I would use The Ratio Test (or L'Hopital's) on this:

    Let a_n=L^n

    \frac{|a_{n+1}|}{|a_n|}=\frac{L^{n+1}}{L^n}

    \frac{|a_{n+1}|}{|a_n|}=\frac{L^n L}{L^n}

    \frac{|a_{n+1}|}{|a_n|}=L

    Since L is between 0 and 1 ( 0 \leq L \leq 1) then the series is convergent.

    The statement is true!

    Thanks again Moo.
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    |a_2-a_1|\lim_{N \rightarrow \infty}\sum_{n=0}^{N-1}L^n

    I would use The Ratio Test (or L'Hopital's) on this:
    Why not just use facts about geometric series?
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  5. #5
    Super Member Showcase_22's Avatar
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    umm, what facts?

    I know that I would be adding increasingly smaller terms each time, but the same goes for \sum_{k=1}^{n} \frac{1}{k} and that diverges.

    So how would you do it?
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  6. #6
    Junior Member Ziaris's Avatar
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    The harmonic series is not a geometric series. It's easy to confuse zeta-type sums with geometric series since they are superficially similar, but note the difference between

    \lim_{A\to\infty}\sum_{n=1}^{A}n^{-s}=1+2^{-s}+3^{-s}+4^{-s}+\cdots and \lim_{A\to\infty}\sum_{n=0}^{A}{k^n}=1+k+k^2+k^3+\  cdots

    Note how in the first sum the number is changing and in the second the number k is constant while the power is changing. Both series in this case are convergent if s>1 for the left, and for the geometric series on the right |k|<1, k\neq 0. Since this is the case, you can conclude that given your work that your series is indeed convergent if 0 < L < 1. (Note that the harmonic series with a finite upper bound is convergent, so that the series you posted \sum_{k=1}^{n}\frac{1}{k} does not \to\infty if n does not \to\infty. )
    Last edited by Ziaris; December 22nd 2008 at 09:39 AM.
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  7. #7
    Super Member Showcase_22's Avatar
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    I knew that the harmonic series wasn't geometric. I was using it as an example (admittedly a bad one! =S) to try and show that a series can tend to infinity even though successive terms get smaller.

    I also apologise for miswriting my sum (I was on a library computer and had to write quickly):

    \sum_{k=1}^{\infty} \frac{1}{k} \rightarrow \infty

    I think that's better
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