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Isn't this weird? (convergence of a sequence)

  1. Oct 15, 2004 #1

    quasar987

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    In order to prove, using the [itex]\epsilon-N[/itex] definition, that the sequence

    [tex]\left\{\frac{n^2}{n^2+4}\right\} = \left\{\frac{1}{1+\frac{4}{n^2}}\right\}[/tex]

    converges towards 1, I did the following: The sequence converges towards 1 if

    [tex]\forall \epsilon>0, \exists N\in\mathbb{R} \ \mbox{such that} \ \forall n\in\mathbb{N}, n>N \Longrightarrow \left|\frac{1}{1+\frac{4}{n^2}}-1\right|<\epsilon[/tex]

    We have that

    [tex]\left|\frac{1}{1+\frac{4}{n^2}}-1\right|<\left|\frac{1}{1+\frac{4}{n^2}}\right|[/tex]

    so if we can find the N for the creature on the right side of the inequality, it will also be true for the one on the left side. And then we solve and I spare you the following steps because my question is that if my line of reasoning is correct so far, then proving that there exist an N for the right member proves that there exists an N for

    [tex]\left|\frac{1}{1+\frac{4}{n^2}}-a\right|[/tex]

    where a in any positive real. So it doesn't REALLY show that the limit is 1 because should I try to prove that the limit is "a, any positive real", I would arrive to the same answer! All it shows is that the sequence converges. Right?
     
    Last edited: Oct 15, 2004
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  3. Oct 15, 2004 #2

    shmoe

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    Actually, we don't. n=1 fails. For large enough n (3 or more), this inequality will be true.

    However, you won't be able to bound the right hand thing above by an arbitrary [tex]\epsilon>0[/tex]. Forget an arbitrary one, take [tex]\epsilon =1/5[/tex] and you won't be able to do it, since it's false for all n (any epsilon less than 1 will also fail).This thing starts at 1/5 and approaches 1, it's not getting arbitrarily small.
     
  4. Oct 15, 2004 #3

    Gokul43201

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    But can you ? Notice that this creature doesn't get smaller than 1/5....ever.

    PS : shmoe's post wasn't up when I started...so ignore this.
     
    Last edited: Oct 15, 2004
  5. Oct 15, 2004 #4

    quasar987

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    Ok, thanks for pointing that out. I start again.

    The sequence converges towards 1 if

    [tex]\forall \epsilon>0, \exists N\in\mathbb{R} \ \mbox{such that} \ \forall n\in\mathbb{N}, n>N \Longrightarrow \left|\frac{n^2}{n^2+4}-1\right|<\epsilon[/tex]

    We have that (with some luck),

    [tex]\left|\frac{n^2}{n^2+4}-1\right|<\left|n^2-1\right|<\left|n^2\right|=n^2[/tex]

    which is an easily tamed creature that produces an N that is true for all N such that

    [tex]N\geq \frac{1}{\sqrt{\epsilon}}[/tex]

    So my argument remains, that is to say, isn't it weird that since showing that this converges to 1 is the same as showing that it converges to any a, positive real. So it doesn't REALLY show that the limit is 1 because should I try to prove that the limit is "a, any positive real", I would arrive to the same answer! All it shows is that the sequence converges. Right?
     
  6. Oct 15, 2004 #5

    Gokul43201

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    It's going to take a lot more than "some" luck !

    Try n=1 and see what happens.

    And I suggest you don't spend a whole lot of time trying more of these ...any such attempt must fail.
     
    Last edited: Oct 15, 2004
  7. Oct 15, 2004 #6

    quasar987

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    Bad luck!

    I found another way of doing it (thanks to Fredrik's trick) that DOESN'T "cause" the paradox I was invoking, so it should be right this time! (notice I used the word "should" :wink:). Here we go again.

    We have that, *crossing my fingers*

    [tex]\left|\frac{n^2}{n^2+4}-1\right|=\left|\frac{n^2+4-4}{n^2+4}-1\right|=\left|1-\frac{4}{n^2+4}-1\right|=\left|-\frac{4}{n^2+4}\right|=\frac{4}{n^2+4}<\frac{4}{n^2}<\epsilon[/tex]
    [tex] \Leftrightarrow n^2>\frac{4}{\epsilon}\Leftrightarrow n>\sqrt{\frac{4}{\epsilon}}[/tex]

    we discarded the option that n was greater than minus the square root because it is trivial by n being a positive interger. So any N satisfying

    [tex]N\geq \frac{2}{\sqrt{\epsilon}}[/tex]

    is fine.
     
    Last edited: Oct 15, 2004
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