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 {AnBn} is bounded but not convergent. Also: Give an example of two sequences tending to infinity such that {AnBn} 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. 
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 AnBn will tend towards infinity and therefore will be unbounded Is this true? Thanks! 
February 13th, 2017, 12:23 PM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,910 Thanks: 774 Math Focus: Wibbly wobbly timeywimey stuff.  
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 AnBn>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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