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"super convergence" proof

  1. Oct 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose (## a_n ##) is a sequence and let [itex]l\in\mathbb R[/itex]. Let us say that (## a_n ##) is "super convergent" to ##l## if there exists [itex]N\in\mathbb N[/itex] such that for every ε>0 we have ##n \geq N## ⇒ |(## a_n - l##|<ε . Show that if (## a_n ##) super converges to l then (## a_n ##) converges to ##l## in the usual sense. What about the converse?


    2. Relevant equations
    The usual definition of convergence as given in my textbook is:

    Let (## a_n ##) be a sequence and let [itex]l\in\mathbb R[/itex]. Then (## a_n ##) converges to ##l## if for every ε>0 there exists [itex]N(ε)\in\mathbb N[/itex] such that ##n \geq N(ε)## ⇒ |(## a_n - l##|<ε

    3. The attempt at a solution

    The only difference I see between the super convergent definition and usual definition is that for super convergence N does not depend on ε as it does in the usual definition. Therefore no matter how small an ε is chosen, any ##n \geq N## will imply (## a_n - l##|<ε . So even if ##N## is dependant on ε as it is in the usual definition, any N will work so the dependence does not matter. Therefore super convergence implies convergence.


    The second part about the converse is what I am confused about. I'm pretty sure that for converse I'm supposed to show that if (## a_n ##) converges in the usual sense, that it also must "super converge" which is obviously not true because if N is dependent on ε for regular convergence, a random N cannot be selected to have the super convergence statement hold. I'm pretty sure I'm on the right track with this question I just don't know how i would go about showing this last part.
     
  2. jcsd
  3. Oct 20, 2014 #2

    Dick

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    Yes, the converse is obviously not true. If you think about what 'super convergence' means, it means the sequence must be constant for n>N. There are sequences that converge that are not eventually constant. Use that for a counterexample. That proves the converse is not true.
     
  4. Oct 21, 2014 #3
    Makes sense thanks.
     
  5. Oct 21, 2014 #4

    PeroK

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    You should note and try to prove the following:

    If ##\forall \ \epsilon > 0, \ \ |x| < \epsilon## then ##x = 0##
     
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