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Bounded sequence as convergent

  1. Jan 16, 2006 #1
    Some rule says that not all bounded sequence must be convergent sequence , one example is the sequence with general bound:
    Xn=(-1)^n

    could anyone help?!
    thanks in advance!
     
  2. jcsd
  3. Jan 16, 2006 #2
    What is it that you want help with? To show that x_n is divergent, it suffices to find two subsequences of x_n which converge to different numbers.
     
    Last edited: Jan 16, 2006
  4. Jan 16, 2006 #3

    matt grime

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    Ooh, erm, let me think... how about looking at the bloody definitions? Yes, I am tired, so delete as applicable, but quite frankly, I've seen enough of this for one day.
     
  5. Jan 16, 2006 #4
    if a sequence converges then it's bounded, but if it's bounded it might not converge as that example shows.
     
  6. Jan 17, 2006 #5
    you're right, the question is how to prove that,could you help?!
     
  7. Jan 17, 2006 #6

    matt grime

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    What is the negation of: x_n converges to x? Show that this is satisfied. Of course if x_n converges to x then all subsequences of x_n converge to x as well, maknig for an easy proof that it doesn't converge. If you're not used to working out the negations of propositions then say so.
     
  8. Jan 17, 2006 #7
    suppose that the sequence {a_n} converges. then a_n --> L for some finite L. now look at the definition of convergence & see that for ANY epsilon (yadda yadda)... for n>N but to make things easier just choose epsilon=1 & if n>N get 2 inequalities |a_n| < |a_n - L| + |L| < 1 + |L| (i think that's right, you check it). if n<N then |a_n| < something else.

    so for any n>0, |a_n| < M where M=max{a certain set of numbers}. end of proof
     
    Last edited: Jan 17, 2006
  9. Jan 17, 2006 #8

    HallsofIvy

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    Prove what? That there exist bounded sequences that are not convergent? You do that by exhibiting one, just as you did.

    Prove that that particular sequence is not convergent? Just as Muzza said in the very first response to your post: show that there exist two different subsequences that converge to two different limits- in this case the subsequence with n even: 1, 1, 1, ... converges to 1, the subsequence with n odd: -1, -1, -1, ... converges to -1. Since arbitrarily far into the sequnce there exist numbers arbitrarily close to 1 and numbers arbitrarily close to -1, taking [itex]\epsilon= 1/3[/itex] will show that no number can be the limit of the entire sequence.
     
    Last edited: Jan 18, 2006
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