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Convergence of a sequence.

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Verify, using the definition of convergence of a sequence, that
    the following sequences converge to the proposed limit.
    a) [itex] lim \frac{1}{6n^2+1}=0 [/itex]
    b) [itex] lim \frac{3n+1}{2n+5}=\frac{3}{2} [/itex]
    c) [itex] lim \frac{2}{\sqrt{n+3}} = 0 [/itex]
    3. The attempt at a solution
    A sequence [itex] a_n [/itex] converges to a real number a if for every ε there is
    an N in the naturals such that whenever n≥N it follows that
    [itex] |a_n-a|< \epsilon [/itex].
    so for the first one I need [itex] \frac{1}{6n^2+1}< \epsilon [/itex]
    and then I turn it into [itex] \frac{1}{\epsilon}<6n^2+1 [/itex]
    So i could pick an n large enough to make that happen.
    on the second one I move the 3/2 over and then combine those
    fractions with a common denominator and I get
    [itex] |\frac{-12}{4n+10}|< \epsilon [/itex]
    Am I doing this right or am I way off.
     
  2. jcsd
  3. Feb 6, 2012 #2

    vela

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    Good start, though the -12 should be -13. Now what?
     
  4. Feb 6, 2012 #3
    so I have [itex] |\frac{-13}{4n+10}<\epsilon | [/itex]
    for any epsilon I can pick an n large enough to make that true.
     
  5. Feb 6, 2012 #4

    vela

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    You should explicitly find what N will work.
     
  6. Feb 6, 2012 #5
    that seems weird to me because [itex] \epsilon [/itex] could be anything
    so how would any fixed N work.
     
  7. Feb 6, 2012 #6

    vela

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    You're given some ##\epsilon > 0##, and then you have to find some N for that given ##\epsilon## such that the implication holds, so N will generally depend on ##\epsilon##.

    For example, if you had
    $$\frac{1}{n} < \epsilon \qquad \Rightarrow \qquad n > \frac{1}{\epsilon},$$ you could choose N to be any integer greater than ##1/\epsilon##.
     
  8. Feb 6, 2012 #7
    so then I just solve for n in terms of [itex] \epsilon [/itex]
    [itex] .25(\frac{-13}{\epsilon }-10)<n [/itex]
    thanks for your help by the way
     
  9. Feb 6, 2012 #8

    vela

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    Since you're working with inequalities, you can do some simplifications to make the algebra less tedious:
    $$\left|\frac{-13}{4n+10}\right| < \frac{16}{4n} = \frac{4}{n} < \epsilon$$So instead of that complicated expression you have, you can choose N to be an integer greater than 4/ε.
     
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