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Math Help - Epsilon-limit proof for real number sequences

  1. #1
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    Epsilon-limit proof for real number sequences




    [note: also under discussion in s.o.s. math board]
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    This is very well known question. It is generally known as the Sequence of Means.

    It is done in two cases. First prove it for the case L=0.

    Then if L\ne 0 define b_n=a_n-L and use the first case.
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    I don't see how to work out the epsilon-limit proof.
    At the end, we have to show that:
    for all ε>0, there exists an integer M such that
    n≥M => |[(a1+a2+...+an)/n] - L|< ε.

    But how can we find that M?

    (even for the case L=0 I am facing the same trouble...I can't find a way to link |[(a1+a2+...+an)/n] - L| with |a_n -L|, so I don't know how to find M)

    Any help is appreciated!
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    Have you had that “intensive one-on-one instruction/tutorial” that I suggested for you earlier?
    I seems to me that you have some real issues with this material.

    You need help beyond which you can expect from websites.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Have you had that “intensive one-on-one instruction/tutorial” that I suggested for you earlier?
    I seems to me that you have some real issues with this material.

    You need help beyond which you can expect from websites.
    I would LOVE to have it, but unfortunately I don't have the fortune to have such a treasury experience...I'm learning it from my textbooks.

    But could you kindly point out which part I was wrong that leads you to make the above comment? I don't think I've said anything conceptually wrong yet...
    We have to show that (by definition of convergence):
    for all ε>0, there exists an integer M such that
    n≥M => |[(a1+a2+...+an)/n] - L|< ε.

    and just like every epsilon-limit proof of this type, we have to actually find a specific M that works, right? And to do this, I believe we need to link |[(a1+a2+...+an)/n] - L| with |a_n -L|. And right now, I am stuck on this step. I know that the triangle inequality may help, but I am not sure how to apply it in this case.

    Could someone help me, please?
    Any help is much appreciated!
    Last edited by kingwinner; January 18th 2010 at 05:18 PM.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kingwinner View Post
    I would LOVE to have it, but I don't have the fortune to have such a treasury experience...I'm learning it from my textbooks.

    But could you kindly point out which part I was wrong that leads you to make the above comment? I don't think I've said anything conceptually wrong yet...
    We have to show that (by definition of convergence):
    for all ε>0, there exists an integer M such that
    n≥M => |[(a1+a2+...+an)/n] - L|< ε.

    and just like every epsilon-limit proof of this type, we have to actually find a specific M that works, right? And to do this, I believe we need to link |[(a1+a2+...+an)/n] - L| with |a_n -L|. And right now, I am stuck on this step. I know that the triangle inequality may help, but I am not sure how to apply it in this case.

    Could someone help me, please?
    Any help is much appreciated!
    Hey king!

    How about combining the fractions?

    \frac {a_1 + a_2 +  \cdots + a_n}n - L = \frac {a_1 + a_2 + \cdots + a_n - nL}n = \frac {a_1 - L + a_2 - L + \cdots a_n - L}n  = \frac {a_1 - L}n + \frac {a_2 - L}n + \cdots + \frac {a_n - L}n

    see how to apply the triangle inequality now?
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    Quote Originally Posted by Jhevon View Post
    Hey king!

    How about combining the fractions?

    \frac {a_1 + a_2 + \cdots + a_n}n - L = \frac {a_1 + a_2 + \cdots + a_n - nL}n = \frac {a_1 - L + a_2 - L + \cdots a_n - L}n  = \frac {a_1 - L}n + \frac {a_2 - L}n + \cdots + \frac {a_n - L}n

    see how to apply the triangle inequality now?
    OK, after using the triangle inequality, I will have n terms.
    |(a_1 -L)/n|, |(a_2 -L)/n|,...,|(a_(n-1) -L)/n|, |(a_n -L)/n|

    How can I bound each of these?
    My idea is to set each of these to be less than ε/? (I don't know what to use for the ? part) and solve the inequality for n, but since it contains terms like a_(n-1) and a_n in the numerator, I don't think I can solve for n?
    I have no clue how to continue, can you please show me how?

    Thanks!
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    OK, after using the triangle inequality, I will have n terms.
    |(a_1 -L)/n|, |(a_2 -L)/n|,...,|(a_(n-1) -L)/n|, |(a_n -L)/n|

    How can I bound each of these?
    My idea is to set each of these to be less than ε/? (I don't know what to use for the ? part) and solve the inequality for n, but since it contains terms like a_(n-1) and a_n in the numerator, I don't think I can solve for n.
    I have no clue how to continue, can you please show me how?

    Thanks!
    Your \varepsilon cannot depend on n.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kingwinner View Post
    OK, after using the triangle inequality, I will have n terms.
    |(a_1 -L)/n|, |(a_2 -L)/n|,...,|(a_(n-1) -L)/n|, |(a_n -L)/n|

    How can I bound each of these?
    My idea is to set each of these to be less than ε/? (I don't know what to use for the ? part) and solve the inequality for n, but since it contains terms like a_(n-1) and a_n in the numerator, I don't think I can solve for n?
    I have no clue how to continue, can you please show me how?

    Thanks!
    By the way, I obliged you, but i wouldn't really use the triangle inequality here, well, not like this at least. finding something to bound each of these expressions seems like too much work, or at least odd. pick your M to be one of the subscripts in a_1,~a_2,\dots,~a_n. Bound the sum of all |a_i - L|/n of everything below it (that is, the expressions for which i \le M), and each term above it could be bounded by some \epsilon. See if you can make something like that work.
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