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Sequence (a_n)^2 ->0 implies (a_n) ->0 ?

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Let an be an infinite sequence.

    1) Is it true that an2 ->0 as n->∞
    => an ->0 as n->∞ ?

    2) Is it also true that an ->0 as n->∞
    => an2 ->0 as n->∞ ?

    2. Relevant equations
    N/A

    3. The attempt at a solution
    Both SEEM to be true to me, but I am not sure why. What is the simplest way to explain these? I want to understand intuitively.

    Any help is appreciated!
     
  2. jcsd
  3. Nov 24, 2009 #2

    HallsofIvy

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    Saying that [itex]a_n\to 0[/itex] means that, given any [itex]\epsilon> 0[/itex] there exist N such that if n> N, [itex]|a_n- 0|= |a_n|< \epsilon[/itex].

    Saying that [itex]a_n^2\to 0[/itex] means that, given any [itex]\epsilon> 0[/itex] there exist N such that if n> N, [itex]|a_n^2- 0|= |a_n^2|< \epsilon[/itex].

    If you know that [itex]a_n\to 0[/itex] then, given any [itex]\epsilon> 0[/itex] there exist N such that if n> N, [itex]|a_n|< \sqrt{\epsilon}[/itex]. From that it follows that, for n> N, [itex]|a_n^2|< \epsilon[/itex].

    Conversely, if you know that [itxex]a_n^2\to 0[/itex] then, given any [itex]\epsilon> 0[/itex], there exist N such that if n> N, [itex]|a_n^2|< \epsilon^2[/itex]. From that it follows that, for n> N, [itex]|a_n|< \epsilon[/itex].

    More generally, if [itex]a_n\to a[/itex] and f(x) is continuous in some neighborhood of a, then [itex]f(a_n)\to f(a)[/itex].
     
  4. Nov 24, 2009 #3
    Thanks! But I have a question, then.
    an2 ->0 as n->∞
    => an ->0 as n->∞

    But the square root function is NOT continuous at 0, and it's definitely NOT continuous in a neighbourhood of 0, so the theorem does not apply??
    It's weird...
     
    Last edited: Nov 24, 2009
  5. Nov 24, 2009 #4

    Dick

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    Halls said 'more generally'. sqrt is one-sided continuous at 0, from the positive side. If you don't have special theorems about one sided continuity, then just pay attention to his suggestions about an epsilon delta proof. Isn't that what you meant about "SEEM to be true"?
     
  6. Nov 24, 2009 #5
    I have a question about the proof.

    Saying that [itex]a_n\to 0[/itex] means that, given ANY [itex]\epsilon> 0[/itex] there exist N such that if n> N, [itex]|a_n- 0|= |a_n|< \epsilon[/itex].

    Because of the word "ANY", we can take epsilon here to be [itex]\sqrt{\epsilon}[/itex], and there would still exist some N, right?
     
    Last edited: Nov 24, 2009
  7. Nov 24, 2009 #6

    Dick

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    You can if you know as epsilon approaches zero, sqrt(epsilon) approaches zero. If you are ok with that, go for it.
     
    Last edited: Nov 24, 2009
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