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February 13th, 2017, 11:20 AM   #1
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sequence question:

Give an example of two sequences tending to infinity such that {An-Bn} is bounded but not convergent.

Also:

Give an example of two sequences tending to infinity such that {An-Bn} is unbounded.

Last edited by skipjack; February 13th, 2017 at 11:36 AM.
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February 13th, 2017, 11:39 AM   #2
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An = n, Bn = n - (-1)^n
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February 13th, 2017, 11:42 AM   #3
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Is this for part A? Thanks.

Last edited by skipjack; February 13th, 2017 at 11:54 AM.
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February 13th, 2017, 11:44 AM   #4
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Also, would it be correct to say the sequence you gave me would be bounded within the integers?

Last edited by skipjack; February 13th, 2017 at 11:54 AM.
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February 13th, 2017, 11:59 AM   #5
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I gave an answer for your first question. The second question is easy.

I'm not sure what you mean by "bounded within the integers", but (-1)^n is clearly bounded.

Throughout, I assumed that your n is given the values 1, 2, 3, etc., to generate the sequences.
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February 13th, 2017, 12:12 PM   #6
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for the second question could you have An=n! and Bn=2^n
Since N! "Beats" 2^n to infinity then surely An-Bn will tend towards infinity and therefore will be unbounded

Is this true? Thanks!
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February 13th, 2017, 12:23 PM   #7
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Quote:
Originally Posted by Jaket1 View Post
for the second question could you have An=n! and Bn=2^n
Since N! "Beats" 2^n to infinity then surely An-Bn will tend towards infinity and therefore will be unbounded

Is this true? Thanks!
It's a good hint. Let's see some work...how would you prove this?

-Dan
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February 13th, 2017, 12:27 PM   #8
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For this too work then for any K>0, then there exists an N such that for all n>N then |An-Bn|>K

Would this be appropriate working out?
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February 13th, 2017, 12:37 PM   #9
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or if you want to prove that n! beats 2^n im guessing you could use induction? Would this work?
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