Convergence Proof for xn/xn+1: Need Help!

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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?
 
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It's not true. Try to find a counterexample.
 
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
 
bballninja said:
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.
 
Well would I be able to claim that it is non-decreasing monotone, and show that it is bounded which implies convergence?
 
bballninja said:
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?
 
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
 
bballninja said:
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.
 
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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