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Homework Help: Did i prooved this convergense corretly

  1. Dec 1, 2008 #1
    i was needed to prove the convergence of the following sequence

    http://img233.imageshack.us/img233/2353/img9143qb3.jpg [Broken]

    is it correct?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 1, 2008 #2


    Staff: Mentor

    You can check that it works. For example, if epsilon = .01 is given, what is N? And will (n - 1)/n^2 be less than epsilon?
  4. Dec 1, 2008 #3
    if e=0.1
    means n>10


    so its correct
    but does my way of solving is correct regarding formality??
  5. Dec 1, 2008 #4


    Staff: Mentor

    Your inequality above uses N = 20, not N = 11. Does it work for N = 11?

    As far as your proof is concerned, you have a lot of extra stuff in it that you don't need, but there are some things that you don't have that you should have.

    The idea of this type of proof is that, given any positive number epsilon, you tell the reader a number N, so that for all n > N, |f(x) - L| < epsilon. After you have solved for n in terms of epsilon, you should say "Take N = <whatever you found>"
  6. Dec 2, 2008 #5
    N is the smallest number of steps for which |An-0|<e
    n should be bigger n>N

    why N=20

    N should be smaller then N
  7. Dec 2, 2008 #6
    The only mistake i can see is that you haven't fixed epsilon to take strictly positive arguments.
    Suppose you took a negative value..then your in-equality wouldnt hold.
  8. Dec 2, 2008 #7
    so for e=0.1


    whats the possible values for N?

    as i can see it can take any of this values N=0 ,1,2..10
  9. Dec 2, 2008 #8


    Staff: Mentor

    That was my question. You said N = 11, but then used 20.
    N can't possibly be smaller than N
  10. Dec 2, 2008 #9
    i meant
    n can't possibly be smaller than N
  11. Dec 2, 2008 #10


    Staff: Mentor

    I don't think you're getting it. Think of the proof as a dialog between two people.

    You: lim (n - 1)/n^2 = 0, as n approaches infinity.
    Me: Oh, yeah? Prove it.
    You: I can make (n -1)/n^2 as close to zero as you like.
    Me: OK, can you get it smaller than 1/10.
    You: Yes. Take N = 10. a_11 = 10/11^2 = 10/121 < 0.1. And all of the other terms in the sequence, a_12, a_13, etc. are even smaller. Are you satisfied?
    Me: Well, 1/10 isn't really that small. Can you make (n - 1)/n^2 smaller than .001?
    You: Sure, take N = ...

    Do you see how this works? Someone supplies an epsilon, and you supply a number N, that is calculated from that epsilon.
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