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Help with a convergence proof

  • Thread starter bballninja
  • Start date
  • #1

Homework Statement



If xn-> ∞ then xn/xn+1 converges.

Homework Equations





The Attempt at a Solution



I can see why the statement is true intuitively, but do not know how to make a rigorous proof. I have looked at the definitions of divergence/convergence but can get any ideas of how to prove this. Do I maybe start by showing that xn/xn+1 is bounded?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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It's not true. Try to find a counterexample.
 
  • #3
I'm pretty sure it's true, since each successive term of the sequence will be larger or equal to the previous, so xn/xn+1 should always be ≤ 1
 
  • #4
Dick
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I'm pretty sure it's true, since each successive term of the sequence will be larger or equal to the previous, so xn/xn+1 should always be ≤ 1
That would be true if the convergence were monotone (i.e. xn is increasing). But even if it were, that wouldn't prove it converges. There are a lot of numbers between 0 and 1.
 
  • #5
Well would I be able to claim that it is non-decreasing monotone, and show that it is bounded which implies convergence?
 
  • #6
Dick
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Well would I be able to claim that it is non-decreasing monotone, and show that it is bounded which implies convergence?
(1,1,2,2,4,4,8,8,16,16,...). Does it converge to infinity? What about your ratio?
 
  • #7
Ahh I'm so sorry haha. The instructor just emailed us that there was a typo and the ratio should actually be xn / (xn+1). This makes more sense now. Thanks for your help though
 
  • #8
Dick
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Ahh I'm so sorry haha. The instructor just emailed us that there was a typo and the ratio should actually be xn / (xn+1). This makes more sense now. Thanks for your help though
No problem, you're welcome. The correction makes a BIG difference.
 
  • #9
I still can't come up with an answer and my presentation is at 10.

So far I've been able to show that since xn→∞, then 1/xn→0. Then

1/(xn+1) < 1/xn < ε

If anyone is available to help me, that would be very appreciated.
 

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