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Sequences - Definition of convergence

  1. Jun 29, 2013 #1
    Alright, I need some help with this.

    an = [itex]\frac{1 - 5n^{4}}{n^{4} + 8n^{3}}[/itex]

    To find the limit of convergence, use l'Hopital's Rule. The result will come out to

    L = -5

    From my book,
    "The sequence {an} converges to the number L if for every positive number ε there corresponds an integer N such that for all n,
    n > N → | an - L | < ε"

    So, to check that L = -5 is true, substitute in? How do I show that L = -5 using this definition?

    | an - L | < ε

    | [itex]\frac{1 - 5n^{4}}{n^{4} + 8n^{3}}[/itex] - (-5) | < ε

    Let n = 1, ε = 1

    | [itex]\frac{1 - 5}{1 + 8}[/itex] + [itex]\frac{45}{9}[/itex]) | < ε = 1

    [itex]\frac{41}{9}[/itex] < ε = 1, which is not true

    Thank you for your time! This definition is very confusing to me for some reason.
     
  2. jcsd
  3. Jun 29, 2013 #2

    WannabeNewton

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    That's not how the definition works. You have to take an arbitrary ##\epsilon > 0## and show that there exists an ##N \in \mathbb{N}## (which may or may not depend on ##\epsilon##) such that for all ##n> N## we have that ##\left | \frac{1 - 5n^{4}}{n^{4} + 8n^{3}} + 5 \right | < \epsilon##.
     
  4. Jun 29, 2013 #3

    mathman

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    The basic idea is that for there is an N (large enough) so that the ε test will hold for all n > N.
     
  5. Jun 29, 2013 #4
    Ohhh, so my e can be any positive number. For convergence, I must find an N.

    thank you!
     
  6. Jun 29, 2013 #5

    WannabeNewton

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    It might help to see a trivial example first. Consider the sequence given by ##a_n = \frac{1}{n}## and let ##\epsilon > 0##. Intuitively we can see that this converges to ##L = 0## but let's show that this is indeed the case. There exists an ##N\in \mathbb{N}## such that ##\epsilon N > 1## i.e. ##\frac{1}{N} < \epsilon##. Thus we have that ##\left | \frac{1}{n} \right | < \epsilon## for all ##n\geq N## meaning ##a_n\rightarrow 0##, as we would expect.
     
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