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A textbook question

  1. Nov 30, 2005 #1
    In my textbook there is an example that shows determination whether a series converges or diverges using the limit comparison test.
    "
    The series is (1+n *ln(n)) / (n^2 + 5)
    For n large, we expect an to behave like (n*ln(n))/n^2 = (ln(n))/n, which is greater than 1/n for n>= 3, so we take bn = 1/n.
    "

    My question is why can I use 1/n as an comparison series, I guess it's because its smaller than (ln(n))/n for large n but I'm not sure how that should validate the choice.

    Could someone please explain to me?
     
  2. jcsd
  3. Nov 30, 2005 #2

    HallsofIvy

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    "I guess it's because its smaller than (ln(n))/n for large n "

    Yes, that's the whole point- together with the fact that [itex]\Sigma \frac{1}{n}[/itex] does not converge. The "Limit Comparison Theorem" says that if [itex]a_n\leq b_n[/itex] for large enough n, and [itex]\Sigma a_n[/itex] does not converge, then [itex]\Sigma b_n[/itex] does not converge.
     
  4. Nov 30, 2005 #3
    they used 1/n because they had a hunch that the series diverges. so when you do the limit comparison test you calculate lim an/bn and if the answer is a finite number greater than zero then either both of your series converge or both diverge. since 1/n is a known divergent series the series you compared it to must also diverge.
    it is good test to know for an exam because it is so easy and doesnt require a bunch of inequalities and lots of thinking. if you suspesct your series an to converge then just pick a series bn that converges and calculate lim an/bn if its finite and bigger than zero then you are done.
     
  5. Nov 30, 2005 #4
    Thanks.:smile:
     
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