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Diverging sequence

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data

    show that the sequence is divergent:
    xn: 1,0,1,0,1,0,.....

    3. The attempt at a solution

    there exists a number p>0 such that there is no N with the property abs(xn-C)<p
    where n>N for all n and C is the limit.

    I don know how to find this value of p. My teacher said that p must be greater than 1/2 but i cant understand why.
    can someone please explain using very basic explanation.

    thank you very much.
  2. jcsd
  3. Oct 11, 2008 #2
    Suppose, for contradiction, that x_n is convergent. Then lim x_n exists, call it C. That is, for any p > 0, we can find N s.t. |x_n - C| < p for all n > N.

    It follows then that |x_n - C| < 1/2 and |x_(n+1) - C| < 1/2, for p = 1/2. Can you continue from here? Hint: if x_n = 0, then x_n+1 = 1. Also, use the triangle inequality to find the contradiction.
    Last edited: Oct 11, 2008
  4. Oct 11, 2008 #3


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    Hi sara_87! :smile:

    uh? your teacher must mean p < 1/2 :confused:

    whatever C you choose, either 0 or 1 will be further away from C than p, if p < 1/2. :smile:
  5. Oct 11, 2008 #4
    Haha yes, noted in my post as well; picking p =1 would yield nothing substantive in the argument.

    Is this for an Analysis class, sara_87?
  6. Oct 11, 2008 #5
    Yes this is for an analysis class.

    sorry, i ment p<1/2.
    Thanx JinM and tiny-tim.
  7. Oct 11, 2008 #6
    Hold on, i dont fully understand yet;
    lets just say p=3 then why doesnt this work because abs(xn-1)<3

    I think im missing the point???????
  8. Oct 11, 2008 #7
    Relax. The sequence diverges, but picking p = 3 won't help you because the limits 1 and 0 are both inside the interval (-3,3), so arguing divergence is useless here. Pick something smaller until one of the limits leave the interval. You're just trying to constrain an infinite number of terms inside a neighborhood, assuming that a limit exist, but the terms keep leaving the interval so there is your contradiction. Just make this more rigorous as I outlined for you above.
  9. Oct 11, 2008 #8


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    Yup … you've started trying to prove that it does converge! :smile:
  10. Oct 11, 2008 #9
    ok so lets take p=2/3; why doesnt this work for:
    abs(xn-c)<p ??
  11. Oct 11, 2008 #10
    Sara, I'm not an expert on this, as I'm taking the class right now.

    It works for p = 2/3; that is, you would leave out the 1 and constrain the 0 inside the interval, but it is not an ideal candidate for a contradiction argument. From the outline above, what you're trying to do is simply argue that

    1 = |1-C+C| <= |1-C| + |C| = |1-C| + |0-C| < p + p = 2p.

    Come up with a p where 2p < 1. This will give you p <= 1/2. But do you see why this is a contradiction? Don't forget, you're assuming the result holds for all p > 0, but we have a problem here.
  12. Oct 11, 2008 #11


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    If it converges then every subsequence converges and to the same value. Can you find two subsequences that both converge but to different values? No need for [tex] \delta [/tex] and [tex] \varepsilon [/tex] then.
  13. Oct 11, 2008 #12
    Oh ok, i see the point now. we cant find a value p which is > 1/2 that would satisfy our property abs(xn-C) so that means there is no limit.

    Thanks all.
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