My Math Forum sequence question:

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 February 13th, 2017, 10:20 AM #1 Member   Joined: Jan 2016 From: Blackpool Posts: 45 Thanks: 1 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 10:36 AM.
 February 13th, 2017, 10:39 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,053 Thanks: 1395 An = n, Bn = n - (-1)^n
 February 13th, 2017, 10:42 AM #3 Member   Joined: Jan 2016 From: Blackpool Posts: 45 Thanks: 1 Is this for part A? Thanks. Last edited by skipjack; February 13th, 2017 at 10:54 AM.
 February 13th, 2017, 10:44 AM #4 Member   Joined: Jan 2016 From: Blackpool Posts: 45 Thanks: 1 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 10:54 AM.
 February 13th, 2017, 10:59 AM #5 Global Moderator   Joined: Dec 2006 Posts: 18,053 Thanks: 1395 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. Thanks from Jaket1
 February 13th, 2017, 11:12 AM #6 Member   Joined: Jan 2016 From: Blackpool Posts: 45 Thanks: 1 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!
February 13th, 2017, 11:23 AM   #7
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Quote:
 Originally Posted by Jaket1 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

 February 13th, 2017, 11:27 AM #8 Member   Joined: Jan 2016 From: Blackpool Posts: 45 Thanks: 1 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?
 February 13th, 2017, 11:37 AM #9 Member   Joined: Jan 2016 From: Blackpool Posts: 45 Thanks: 1 or if you want to prove that n! beats 2^n im guessing you could use induction? Would this work?

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