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Sequence convergence

  • #1

Homework Statement


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.


Homework Equations





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.
 

Answers and Replies

  • #2
Dick
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Do you know why lim n->infinity of c^n goes to 0? Hint: take the log.
 
  • #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
 
  • #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?
 
  • #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
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
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
Sure. That's it.
 

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