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Math Help - Help with sequence counterexample

  1. #1
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    Help with sequence counterexample

    Suppose that a sequence (xn) satisfies the following inequality:
    abs(xn+1 - xn) < abs(xn - xn-1) for n = 2,3,4,.....

    Does this sequence converge or is there a counterexample for it being divergent? If it converges, how would you prove it?

    I would love some help on this particular problem. I'm not really sure if it converges or not. Thanks.
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  2. #2
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    Quote Originally Posted by PvtBillPilgrim View Post
    Suppose that a sequence (xn) satisfies the following inequality:
    abs(xn+1 - xn) < abs(xn - xn-1) for n = 2,3,4,.....

    Does this sequence converge or is there a counterexample for it being divergent? If it converges, how would you prove it?

    I would love some help on this particular problem. I'm not really sure if it converges or not. Thanks.
    Let x_n = \sum_{i=1}^n \frac{1}{i}.
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  3. #3
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    Hello, PvtBillPilgrim!

    Suppose that a sequence x_n satisfies the following inequality:
    . . |x_{n+1} - x_n| \;< \;|x_n - x_{n-1}| . for n \,=\,2,3,4,\cdots

    Does this sequence converge or is there a counterexample for it being divergent?
    If it converges, how would you prove it?

    TPH is absolutely correct.
    . . Let me translate it . . .

    It says: the difference of consecutive terms is decreasing.
    . . The terms are "getting closer."
    It would seem that the series converges.

    But the Harmonic Series: . 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots . satisfies the inequality,
    . . and it diverges.

    We have found a counterexample.


    Therefore, the sequence may or may not converge.

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  4. #4
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    However if |x_{n+1} - x_n | \leq k |x_n - x_{n+1}| where 0<k<1 then it will converge.
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