My Math Forum sequence question:

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 February 13th, 2017, 11:20 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2 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.
 February 13th, 2017, 11:39 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,879 Thanks: 1835 An = n, Bn = n - (-1)^n
 February 13th, 2017, 11:42 AM #3 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2 Is this for part A? Thanks. Last edited by skipjack; February 13th, 2017 at 11:54 AM.
 February 13th, 2017, 11:44 AM #4 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2 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.
 February 13th, 2017, 11:59 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,879 Thanks: 1835 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, 12:12 PM #6 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2 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, 12:23 PM   #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, 12:27 PM #8 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2 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, 12:37 PM #9 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2 or if you want to prove that n! beats 2^n im guessing you could use induction? Would this work?

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