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April 19th, 2017, 02:55 AM  #1 
Member Joined: Apr 2014 From: Greece Posts: 47 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: 6,857 Thanks: 2230 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: 2,524 Thanks: 641 
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: 47 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: 1,079 Thanks: 87 
$\displaystyle a_{n}L<\epsilon$, 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 

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converge, proving, sequence 
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