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Thread: Convergence Sequence Question

  1. #1
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    Convergence Sequence Question

    Q: Suppose that {Xn} converges to X. For each n within the set of Natural Numbers, define Yn = (X1+X2+...+Xn)/n. Prove that {Yn} also converges to x.

    My solution: So far, I can only think of this as a continuous function problem. I set {Yn} as a function of Y(x). So if Y(x) is continuous, then Y(Xn) -> Y(x).

    Am I having the right idea?

    Thank you.

    KK
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Q: Suppose that {Xn} converges to X. For each n within the set of Natural Numbers, define Yn = (X1+X2+...+Xn)/n. Prove that {Yn} also converges to x.
    It is sooo late now, I cannot post a complete proof but this is going to be a mess. Hope it will at least help you somehow. (I probably end up deleting this embarassement tomorrow).

    By definition of sequences converging to X,
    $\displaystyle n\geq N\to |x_n-X|<\epsilon/n $

    Prove that $\displaystyle \{y_n\}$ converges to X.
    That is,
    $\displaystyle n\geq N \to \left| \frac{x_1+x_2+...+x_n}{n}-X \right|<\epsilon$

    But,
    $\displaystyle \left| \frac{x_1+x_2+...+x_n}{n}-X\right|<\epsilon$
    Is,
    $\displaystyle \left| \frac{x_1+x_2+...+x_n-nX}{n} \right|<\epsilon$
    Thus,
    $\displaystyle \left| \frac{(x_1-X)+(x_2-X)+...+(x_n-X)}{n} \right|<\epsilon$
    Triangular inequality, and first inequality
    $\displaystyle |x_1-X|+|x_2-X|+...+|x_n-X|<\epsilon/n+...+\epsilon/n$
    Thus,
    $\displaystyle |x_1-X|+...+|x_n-X|<\epsilon$
    Which is true.
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  3. #3
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    Thanks for the reply. I understand how you sub into the |{Yn} - x| thing, but I don't understand how $\displaystyle |x_1-x|+|x_2 -x|+...+|x_n-x|<\epsilon/n+... $, what I question is how do you know $\displaystyle x_1$ is < $\displaystyle x_n$?

    Would you please explain that part? Then I should be good for the rest.

    Thank you!!!

    KK
    Last edited by tttcomrader; Oct 25th 2006 at 05:16 PM.
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