I have that the sequence [itex]a_n=\{2-(-1)^n\}[/itex] not converges.

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  • #1
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I have that the sequence [itex]a_n=\{2-(-1)^n\}[/itex] not converges. I must show this with the rigorous definition.

I think use [itex]\exists{\epsilon>0}\forall{N\in\mathbb{N}}\exists{n\geq N}:|a_n-\ell|\geq\epsilon[/itex]

How i can continue?
 

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  • #2
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Try showing that it's not a cauchy sequence instead and then just say "therefore it is not convergent."
 
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  • #3
HallsofIvy
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Alternatively, note that [itex]a_n= 1[/itex] for n even, [itex]a_n= 3[/itex] for n odd. For any l, there exist arbitarily large n such that [itex]|a_n- l|> 1[/itex], half the distance between 1 and 3.
 
  • #4
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Alternatively, note that [itex]a_n= 1[/itex] for n even, [itex]a_n= 3[/itex] for n odd. For any l, there exist arbitarily large n such that [itex]|a_n- l|> 1[/itex], half the distance between 1 and 3.
So which do you think should be the value of ε>ο ??
 
  • #5
Deveno
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So which do you think should be the value of ε>ο ??
HallsofIvy just told you what epsilon to use, half the value of the difference of the two possible values any term of the sequence can have.
 

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