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Sequence convergence

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose {xn} is a sequence of positive numbers for which there exists c, 0<c<1, such that ([x][/n+1])/([x][/n])<c for all n in Z+. Prove that [x][/n] goes to zero.


    2. Relevant equations



    3. The attempt at a solution
    Let the first term of {xn} be x, where n=1. Then by the given, [x][/n+1]/[x][/n]<1, therefore, [x][/1]>[x][/2]>[x][/3]>...>[x][/n]>[x][/n+1], hence sup{[x][/n]} = [x][/1].
    By the given, inf{[x][/n]}=0 so {xn} is bounded and strictly decreasing. We know a monotone sequence converges if and only if it is bounded, but I am having trouble proving that {xn} goes to zero.
     
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  3. Mar 10, 2010 #2

    Dick

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    Do you know why lim n->infinity of c^n goes to 0? Hint: take the log.
     
  4. Mar 10, 2010 #3
    I understand why lim n -> ∞ c^n = 0, but I don't understand how it relates to the problem. I need to determine why {xn} goes to zero
     
  5. Mar 10, 2010 #4

    Dick

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    x1<c*x0. x2<c*x1<c*(c*x0)=c^2*x0. x3<c*x2<c*(c^2*x0)=c^3*x0. Following?
     
  6. Mar 10, 2010 #5
    since i can show that c^n*x0>xn and since c^n -> 0 then I can use the squeeze theorem to show that {xn} goes to zero...
    just want to make sure I made sense of that correctly
     
  7. Mar 11, 2010 #6

    Dick

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    Sure. That's it.
     
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