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Epsilon- N Proof a Sequence Diverges

  1. Sep 1, 2015 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    1. Let ##x_n = \frac{n^2 - n}{n} ## does ##x_n## converge or diverge?
    2. Let ##x_n = \frac{(-1)^n +1}{n} ## does ##x_n## converge or diverge

    2. Relevant equations
    A sequence converges if ##\forall \epsilon > 0##, ##\exists N \in \mathbb{N} ## such that ##n\geq N ## implies ## | x_n - L | < \epsilon ##
    A sequence diverges if ##\exists \epsilon_{0} > 0##, ##\forall N \in \mathbb{N}##, ##\exists n_{0} \geq N ## such that ##|x_{n_{0}} - L | > \epsilon_{0} ##

    3. The attempt at a solution
    For one, I noticed that ##x_n = n-1 ## which obviously diverges. But as my teacher says, I gotta prove it. So I take ##\epsilon_0 = \frac{1}{2}, \, \, L \in \mathbb{R}, \, \, n \geq N, \, \, N \in \mathbb{N} ##
    $$
    1.) \, \, | (N - 1) - L | < \frac{1}{2}, \, \, \, \, \, \, \, \, |((N+1)-1) - L | = |N-L| <\frac{1}{2}
    $$

    I can apply the triangle inequality to find that ##|N-1-L| - |N-1| \le 1 ##, but I can see from 1.) that ## |N-1-L| - |N-L| < 0 ## Therefore ##x_n## diverges.

    Obviously I wouldn't be posting here if that satisfied me, it doesn't. ##|N-1-L| - |N-L|=(-\infty, \, 0) ## would satisfy both of those.

    2.) I noticed that ##x_n = \frac{(-1)^n +1}{n} > \frac{(-1)^n}{n} ## so all I should have to do is prove ##b_n = \frac{(-1)^n}{n} ## diverges and that should be sufficient. So if we take ##\epsilon_{0} = \frac{1}{2}, \, \, L = 1 \, \, n\geq N ##, ##|b_{n} - 1| > \epsilon ##, ##b_n = 1## when n is even and ##b_n = -1## when n is odd, therefore ##|-1 -1| = |-2| = 2 > \epsilon## which proves ##b_n## is divergent therefore proving ##x_n## is divergent..

    This doesn't satisfy my either, I understanding proving convergence just fine but for some reason I find proving divergence harder.
     
  2. jcsd
  3. Sep 1, 2015 #2

    Geofleur

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    Well, if a sequence diverges, then there will be some ## \epsilon ## such that, no matter how big an ## N ## you pick, ## n \geq N ## will not make ## |x_n - x | < \epsilon ##. That is, we will have ## |x_n - x| \geq \epsilon ## no matter how big ## n ## gets. Does phrasing it that way help clarify things?
     
  4. Sep 1, 2015 #3

    B3NR4Y

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    Not particularly, I understand what the symbols mean, but actually executing a proof of divergence trips me up. I've started over on the first one, but choosing the epsilon and then proving that no matter what, ##|x_n - L| > \epsilon ##, confuses me.
     
  5. Sep 1, 2015 #4

    Geofleur

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    For your first series, you noticed that ## x_n = n - 1 ##. So the first few terms of the series (assuming we start at ## n = 1 ##) are 0, 1, 2, etc. Suppose ## \epsilon ## is, say, 1/2. For the sequence to converge to x, we would need all the ## x_n ## for big enough ## n ## to make | x_n - x | < 1/2. We have the opposite situation - if you make ## n ## big enough, | x_n - x | = | n - 1 - x | will get bigger than any finite number you choose!
     
  6. Sep 1, 2015 #5

    B3NR4Y

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    Oh! Okay, that makes a lot of sense.

    For the second one, if we choose epsilon to equal 1

    $$
    |\frac{(-1)^{n} +1}{n} - L| < 1
    $$
    If n is odd, ## \frac{L}{n} < 1 ##, which isn't necessarily true for all n. L has to be chosen so that ##\frac{1}{L} > \frac{1}{n}##, but L can't be conditioned by n, it has to be fixed. So it's not looking good for ##x_n##

    If n is even, ## \frac{|Ln-2|}{n} < 1##, which isn't true, since Ln - 2 ≥ n, so the fraction ## \frac{|Ln-2|}{n}\geq 1## Therefore ##x_n## diverges.
     
  7. Sep 1, 2015 #6

    Geofleur

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    Write down a bunch of terms of the sequence. Does it look like they are approaching a definite number?
     
  8. Sep 1, 2015 #7

    B3NR4Y

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    For the second one, no, it's either 0 or 1/n ∀n

    But sadly just writing terms isn't sufficient for my professor.
     
  9. Sep 1, 2015 #8

    Geofleur

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    What happens to 2/n when n gets bigger and bigger?
     
  10. Sep 1, 2015 #9

    B3NR4Y

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    It also goes to zero, oops. So it converges and I have to prove that.
     
  11. Sep 1, 2015 #10

    Geofleur

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    You can't prove something that is false, after all :-) That's why it's good to actually write some concrete numbers and get a feel first.
     
  12. Sep 1, 2015 #11

    B3NR4Y

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    Thank you so much! I understand it better now, for sure. :)
     
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