
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 19th, 2017, 02:55 AM  #1 
Member Joined: Apr 2014 From: Greece Posts: 52 Thanks: 0  Proving that a sequence does not converge
I want to prove using the epsilonn theorem that the sequence $\displaystyle b_{n}=(1)^{n}$ does not converge. I know I should use proof by contradiction but it's not clear how to do so. I want to know what is the correct format to write the proof, not the concept behind it. Can anyone please write the analytical proof like how someone should prove it in an exam? I would be grateful 
April 19th, 2017, 03:52 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,939 Thanks: 2265 Math Focus: Mainly analysis and algebra 
You write it and I'll provide feedback. Hint: it is sufficient to demonstrate a value of epsilon for which we can't find a value of n.

April 19th, 2017, 04:03 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,649 Thanks: 680 
You can use the fact that this sequence has two subsequences that converge to 1 and 1 to give you an idea what values to use.

April 19th, 2017, 04:04 AM  #4 
Member Joined: Apr 2014 From: Greece Posts: 52 Thanks: 0 
I know the concept behind it. And I also understand that all I have to do is to find a value for ε so that bnb>= ε What I don't know is how to write it in a mathematically correct way. That's why I asked for the solution so that I can have a basic format to work with on other proofs like this 
April 19th, 2017, 04:10 AM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88 
$\displaystyle a_{n}L<\epsilon$, n>N $\displaystyle \epsilon<a_{n}L<\epsilon$ $\displaystyle L\epsilon<(1)^{n}<L+\epsilon$ $\displaystyle L\epsilon \geq 0$ can't be satisfied. $\displaystyle L\epsilon < 0 \rightarrow L+\epsilon < 2\epsilon$ can't be satisfied 

Tags 
converge, proving, sequence 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
does the following sequence converge or not?  srecko  Calculus  14  November 5th, 2016 11:40 AM 
Sequence does not converge to limit a  illyo  Real Analysis  1  March 15th, 2015 05:50 AM 
proving a property of the fibonacci sequence  juarez.asf  Number Theory  14  April 29th, 2014 02:59 AM 
proving if a sequence is bounded..  nappysnake  Calculus  6  November 28th, 2011 01:39 PM 
Does the Sequence Converge or Diverge?  veronicak5678  Calculus  2  November 4th, 2008 09:18 AM 