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

  1. Dec 21, 2011 #1
    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?
     
  2. jcsd
  3. Dec 21, 2011 #2
    Re: convergence

    Try showing that it's not a cauchy sequence instead and then just say "therefore it is not convergent."
     
    Last edited: Dec 21, 2011
  4. Dec 21, 2011 #3

    HallsofIvy

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    Re: convergence

    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.
     
  5. Dec 21, 2011 #4
    Re: convergence

    So which do you think should be the value of ε>ο ??
     
  6. Dec 21, 2011 #5

    Deveno

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    Re: convergence

    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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