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April 19th, 2017, 02:55 AM   #1
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Proving that a sequence does not converge

I want to prove using the epsilon-n 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
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April 19th, 2017, 03:52 AM   #2
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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.
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April 19th, 2017, 04:03 AM   #3
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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.
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April 19th, 2017, 04:04 AM   #4
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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 |bn-b|>= 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
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April 19th, 2017, 04:10 AM   #5
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$\displaystyle |a_{n}-L|<\epsilon$, 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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