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Math Help - Sequence Convergence proof maybe?

  1. #1
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    Sequence Convergence proof maybe?

    a) Suppose that {X_n} and {Y_n} converge to the same point. Prove that {X_n} - {Y_n} \rightarrow 0 as n \rightarrow \infty.

    b) Show the converse is false.


    Again as usual... some kind of explanation to help me understand this helps alot...
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  2. #2
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    For part a) i would use the \epsilon-\delta definition of limits and it should follow directly. For part b) you just need to show a counter example. Maybe look at y=x\mbox{ and }y^2-x^2=1.
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    ok that was how I started it... thanks for the info...
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    Quote Originally Posted by Caity View Post
    a) Suppose that {X_n} and {Y_n} converge to the same point. Prove that {X_n} - {Y_n} \rightarrow 0 as n \rightarrow \infty.
    b) Show the converse is false.
    Suppose that L is the common limit of the two sequences.
    If \varepsilon  > 0 \Rightarrow \quad \left( {\exists N} \right)\left[ {n \geqslant N \Rightarrow \left| {x_n  - L} \right| < \frac{\varepsilon }<br />
{2}\,\& \,\left| {y_n  - L} \right| < \frac{\varepsilon }<br />
{2}} \right].
    Then observe that \left| {x_n  - y_n } \right| \leqslant \left| {x_n  - L} \right| + \left| {L - y_n } \right|.
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    I finished part a). Thanks for the help. Now about part b). First what exactly would the converse be? I know its If X then Y becomes If Y then X. I was thinking maybe by showing something like subtraction is not commutative. Like if X - Y \rightarrow 0 is not the same as Y - X \rightarrow 0.
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  6. #6
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    The converse is: Suppose that \left( {x_n  - y_n } \right) \to 0 the \left( {x_n } \right) & \left( {y_n } \right) have the same limit.
    Last edited by Plato; October 27th 2008 at 01:25 PM.
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  7. #7
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    Quote Originally Posted by Plato View Post
    The converse is: Suppose that \left( {x_n - y_n } \right) \to 0 the \left( {x_n } \right) & \left( {y_n } \right) have the same limit.

    Well I tried this Plato... but it comes out true and I'm supposed to be proving it false... If the limit of x_n and y_n go to the same number say 1. Then we have 1 - 1 = 0 which shows it is true.
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  8. #8
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    Try this counterexample.
    x_n  = \left( { - 1} \right)^n  + \frac{1}<br />
{n}\,\& \,y_n  = \left( { - 1} \right)^n  - \frac{1}<br />
{n}\,

    Note: we are not given that the sequences converge, only that their difference is null.
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  9. #9
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    Quote Originally Posted by Plato View Post
    Note: we are not given that the sequences converge, only that their difference is null.
    Is it possible to find a counterexample with the two converging ?
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  10. #10
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    Quote Originally Posted by Moo View Post
    Is it possible to find a counterexample with the two converging ?
    The answer is no, the limit must be the same for both sequences.
    Suppose that \left( {x_n } \right) \to L,\quad \left( {y_n } \right) \to K\,\& \,\left| {\left( {x_n } \right) - \left( {y_n } \right)} \right| \to 0.
    That means that almost all the terms of \left( {x_n } \right) are ‘close’ to L; almost all the terms of \left( {y_n } \right) are ‘close’ to K; and almost all the terms of \left( {x_n } \right) - \left( {y_n } \right) are ‘close’ to 0.

    Suppose L \not= K then \varepsilon  = \frac{{\left| {L - K} \right|}}<br />
{4} > 0.
    But how can almost all of the three sequences be with in an \epsilon-distance of its limit?
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