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Convergence proof

  1. Feb 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Show that {n2 - n + 5} is increasing and hence show that {xn} is convergent when

    {xn} = exp[(3n2 - 3n +14) / (n2 - n + 5)]

    You may assume exp x < exp y when x < y, but may not use any properties of the limit of exp x as x → 3.

    2. Relevant equations



    3. The attempt at a solution

    I tried to show xn+1 ≥ xn to show {n2 - n + 5} was increasing.

    So I got (n+1)2 - (n+1) + 5 ≥ n2 - n + 5

    Therefore n2 + n + 5 ≥ n2 - n + 5 and so n ≥ -n

    Does this show that {n2 - n + 5} is increasing? Is it enough proof?

    Going onto the next bit of the question. I assume because you can't use any properties of the limit of exp x as x → 3 that you must use the Sandwich Theorem? This being that when you have sequences such as:

    {zn} ≤ {xn} ≤ {yn}

    that if {zn} and {yn} converge, {xn} also must converge.

    After this I'm not really too sure what to do, or whether this is actually the right way to approach this question.

    Has anyone got any help that they could offer me please?
     
  2. jcsd
  3. Feb 21, 2014 #2

    vela

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    Kinda. Your steps are backwards. You generally shouldn't start with what you're trying to prove. Reverse the steps, starting with something given, like n>0.

    You should be able to show the sequence increases monotonically. Do you have any theorems that apply to such a series to show it converges?
     
  4. Feb 21, 2014 #3
    Thanks for your reply!

    I'm not really sure how to try to show it's increasing the other way around. I find it difficult because it's obvious looking at it that n2 - n + 5 is increasing but proving this in a formal proof is hard.

    I don't think n > 0 necessarily so not sure where to start when reversing the steps like you said but the sequence itself {xn} > 0 as n2 + 5 ≥ n.

    I know about the Monotone Sequence Theorem which is just that if a sequence is increasing/decreasing and is bounded above/below then it is converging. However, I can't see how you can use it in this case because you aren't allowed to use properties of the limit of exp x as x → 3.

    Otherwise you could just divide {xn} = exp[(3n2 - 3n +14) / (n2 - n + 5)] through top and bottom by exp(n2) and it would leave you with a lowest upper bound of e3 and therefore as it's bounded above and is increasing then it must be convergent. I don't think you can do it this way, due to what's stated in the question.
     
  5. Feb 21, 2014 #4

    pasmith

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    You are told to assume that [itex]\exp[/itex] is strictly increasing, so monotone convergence is the correct approach.

    What you can't do is say that [itex]\exp[/itex] is continuous, so that
    [tex]\lim_{n \to \infty} \exp(a_n) = \exp\left(\lim_{n \to \infty} a_n\right)[/tex]
     
    Last edited: Feb 21, 2014
  6. Feb 21, 2014 #5
    wrong topic post sorry
     
  7. Feb 21, 2014 #6
    Thanks for your reply!

    I understand that the Monotone Convergence Theorem requires you to show {xn} is increasing and bounded above. However, I'm not sure how you can show it's bounded above without having to show the limit is exp(3).
     
  8. Feb 21, 2014 #7

    pasmith

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    You just need an upper bound; [itex]e^4[/itex] will suffice.
     
  9. Feb 21, 2014 #8
    Ah right ok so by proving what's inside the brackets for {xn} ≤ 4 will be enough to prove that {xn} is bounded above.

    Then use of the monotonic sequence theorem shows {xn} is converging.

    Thanks for your help!!!!
     
  10. Feb 21, 2014 #9

    vela

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    There has to be some condition on n because if you make n negative enough, ##a_n=n^2-n+5## will be decreasing. n is the index of the sequences, so it's (typically) greater than 0. Starting with n>0, you have:

    n>0
    2n>0
    n>-n

    and then just list the rest of your steps in reverse order.

    The way I would have done it is to calculate what ##a_{n-1}-a_n## is and show the result is positive (with the assumption n>0), from which you can conclude that ##a_{n-1} \gt a_n##.
     
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