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First time poster - Hard limit proof!

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose an and bn are sequences where bn is increasing and approaching positive infinity. Assume that lim n->∞ [ ( an+1 - an ) / ( bn+1 - bn) ]= L, where L is a real number. Prove that lim n->∞ [ an / bn ] = L.

    2. Relevant equations

    Limit theorems

    3. The attempt at a solution

    I tried making an argument with the basis that both sequences have the same number of terms, so L is the limit of (an+1)/(bn+1) and thus also (an/bn)
     
    Last edited: Nov 14, 2012
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  3. Nov 14, 2012 #2

    jbunniii

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    Given [itex]\epsilon > 0[/itex], we can find [itex]N \in \mathbb{N}[/itex] such that
    [tex](L - \epsilon)(b_{n+1} - b_n) \leq a_{n+1}-a_n \leq (L + \epsilon)(b_{n+1} - b_n)[/tex]
    for all [itex]n \geq N[/itex]. This is true by definition of the limit and by the fact that [itex]b_{n+1} > b_n[/itex].

    Big hint: now try summing this inequality starting at [itex]n = N[/itex] up to some arbitrary value, say [itex]n = M[/itex]. Can you conclude anything as [itex]M \rightarrow \infty[/itex]?
     
  4. Nov 15, 2012 #3
    So starting at a certain value N in the sequence, when you take the infinite sum of the inequality an and an+1 would tend to positive infinity along with bn+1 but I can't tie an argument together to actually prove the result.
     
  5. Nov 15, 2012 #4

    HallsofIvy

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    That doesn't make any sense. Both sequences are infinite. That don't have a "number of terms". It is however, true that the limits of the two sequences you mention are the same but that really doesn't matter. The sequence (an+1- an)/(bn+1- bn[/b]) is NOT (an+1/bn+1)- (an/bn).
     
  6. Nov 15, 2012 #5

    jbunniii

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    Try writing out explicitly what you get if you sum the inequality in my previous post from [itex]n = N[/itex] to [itex]n = M-1[/itex]. Notice that all three terms will "telescope." What is the result? Write it out in detail.
     
  7. Nov 15, 2012 #6
    After manipulation were left with (L−ϵ)(1-bN/bM) + aN/bM≤aM/bM≤(L+ϵ)(1-bN/bM) + aN/bM.

    Then would bN/bM converge to zero and we are done?
     
    Last edited: Nov 15, 2012
  8. Nov 15, 2012 #7

    jbunniii

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    Well, you need both [itex]b_N/b_M[/itex] and [itex]a_N/b_M[/itex] to converge to zero as [itex]M \rightarrow \infty[/itex]. Do they?
     
  9. Nov 15, 2012 #8
    I think that's safe to assume without proof since bn is increasing. Thank for your help.
     
  10. Nov 15, 2012 #9

    jbunniii

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    It's not because [itex]b_n[/itex] is increasing. It's because [itex]b_n[/itex] is diverging to positive infinity (a fact you haven't used yet). Therefore [itex]1/b_n[/itex] is converging to...?
     
  11. Nov 15, 2012 #10
    Zero haha. Thanks. xn
     
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