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Prove that a sequence which is bounded above cannot tend to infinity

  1. Oct 15, 2008 #1
    question 1 : Prove that a sequence which is bounded above cannot tend to infinity
    What i did was state the definition ... but I'm trying to proof by contradiction. So i first suppose that a(n) tends to infinity , then a(n) > C . But since it is bounded above , C < or = to U , where U is the upper bound .
    This is where i got stuck. Any ideas ?

    question 2 : I am required to prove the this sequence does not tend to infinity
    B(n) = cos(n^2 + 7)
    how am i suppose to do this ?
     
  2. jcsd
  3. Oct 15, 2008 #2

    HallsofIvy

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    Re: Sequences

    Saying "a(n)> C" makes no sense until you have said what C is!

    Well, just a random guess, but since you were just asked to prove that "if a sequence is bounded above is cannot tend to infinity", how about finding an upper bound for cos(n2+ 7)?:rolleyes: There's an obvious one.
     
  4. Oct 15, 2008 #3
    Re: Sequences

    i got question 2
    does is this statement true : that for any bounded sequence, it cannot tend to infinity ?

    for question 1 , a sequence tend to infinity for every C>0 , there exist a nat no. N such that a(n) > C whenever n>N
    what do you mean i have not said what is C ?
    isn't c just a number ?
     
  5. Oct 15, 2008 #4

    HallsofIvy

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    Re: Sequences

    That is exactly what you were asked to prove in question 1!

    You statement here, "a sequence tends to infinity for every C> 0" makes no sense (the sequence does not depend on any C) and that is NOT what you said in your first post: "Prove that a sequence which is bounded above cannot tend to infinity"- there is no "C" in that.

    My point about not saying what C was is that you said earlier: "So i first suppose that a(n) tends to infinity , then a(n) > C" and THAT makes no sense because you haven't said what C is. You did not say, for example that C was a number. Even if you did say "C is a number" you still wouldn't have proven it. Yes, you must have [itex]C\le U[/itex], but if C were, for example, 2, it might be that the upperbound on {a(n)} was 3! What you meant to say, and should say, is that if {a(n)} "tends to infinity", then for C any number, there exist N such that a(N)> C. If the upperbound on {a(n)} is U, take C= U+ 1. Then what do you have?
     
  6. Oct 15, 2008 #5
    Re: Sequences

    but it reads here from my notes that the definition is stated as i said ... could i then have a correct definition of a sequence that tends to infinity ?
     
  7. Oct 16, 2008 #6

    HallsofIvy

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    Re: Sequences

    What you wrote, and what I was objecting to was:
    " a sequence tend to infinity for every C>0 "
    which should be " a sequence tend to infinity IF for every C>0".
     
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