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Negation of definition of convergence

  1. Oct 26, 2013 #1
    The definition of convergence is given by : ## \forall \epsilon > 0, \exists N \in \mathbb{R} ## such that ## |x_n - l | < \epsilon ## ## \forall n \in \mathbb{N} ## with ## n > N ##

    negate this statement and prove that the sequence ## x_n = (-1)^nn ## is divergent using only the negation of the definition of convergence

    ok here's my attempt of the negation of the statement:

    ## \exists \epsilon > 0 \forall N \in \mathbb{R} ## s.t. ## \exists n \in \mathbb{N} n > N ## and ## |x_n - l | \geq \epsilon ##

    proof that ## x_n = (-1)^nn ## does not converge:

    since the negation of the definition is "## \forall N ##" let N = 1, and find ## n_1 > 1 ## then let ## N = n_1 ## and find ## n_2 > N=n_1 ## etc leaving us with a sequence of indicies ## n_1 < n_2 < n_3.. ##

    so lets prove that it does not converge to 1:

    choose ## epsilon = 1/2 ## then for ## N>0 ## let n be the odd integers i.e. ## n_1 = 1 < n_2 = 3 < n_3 = 5 ... ## then n > N hence ## | (-1)^1(1) - 1 | >= 1/2 ## which is true hence ## x_n## does not converge to 1

    take ## l > 1 ## and choose ##\epsilon ## s.t. ## \epsilon < l - 1 ## again, choosing ## N > 0 ## and n being all the odd integers greater than 0 we have n > N and ## |(-1)(1) - l | \geq \epsilon## which is true

    take ## l < 1## and choose n to be all the odd integers above 0 then n > N and ## |-1 - l | >= \epsilon ## which is true therefore ## x_n ## does not converge

    I think this works, but truth be told, I don't understand the negation of this statement. I mean, I understand HOW I got the negation, but I don't know what it means. The choosing N > 0 is from my lecture notes and I don't understand why that works, and why we can just choose N > 0 if the definition states for all N

    any help please
     
  2. jcsd
  3. Oct 26, 2013 #2
    You are on the right track but I think it is a little simpler. Given the sequence ##a_n = n(-1)^n## you want to show it cannot have a limit. If it does have a limit L then for any ##\epsilon## there is an ##N## such that |##a_n -L##| whenever n > N.

    Choosing ##\epsilon## = 1/2 is a great idea. Now what happens to ##a_n## as n gets large? How does each term relate to the previous one?

    So you want to say something like: no matter what N I choose I can't get |##a_n -L##| < 1/2 for every n > N. Then explain why not.

    The positive statement is I can find an N; the negation is I cannot find an N.
     
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