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Convergence/Divergence and Monotonicity of a Sequence

  1. Mar 20, 2013 #1

    B18

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    1. The problem statement, all variables and given/known data
    Show the following sequence to diverge, or converge. Determine if monotonic.
    a sub n=n+(1/n)

    3. The attempt at a solution
    I understand that the sequence does diverge. I found this because the limit as n→∞ the limit is going to ∞ as well.
    I found that the sequence is monotonic by showing that a sub n is less than a sub n+1. I tried to do the first derivative test for this and was confused when I had the sequence decreasing from 0 to 1 and increasing from 1 to ∞. Wouldn't this make the sequence not monotonic? Just need an explanation to this.
     
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  3. Mar 20, 2013 #2

    LCKurtz

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    Surely you could show ##a_n<a_{n+1}## without resorting to calculus. But the sequence only has values on the integers 1,2,... so it would still be increasing.
     
  4. Mar 20, 2013 #3

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    So we are saying since sequences are only accountable for positive integers meaning the negative result on the first derivative test from 0 to 1 has no effect on monotocity.
     
  5. Mar 20, 2013 #4
    Why did you try the first derivative test on a sequence??
     
  6. Mar 20, 2013 #5

    B18

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    To determine if the sequence is increasing or decreasing to show it is monotonic
     
  7. Mar 20, 2013 #6

    LCKurtz

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    Yes. And I'm also saying that using the derivative test on ##x+\frac 1 x## is way overkill, not to mention the fact that it confused the issue for you.
     
  8. Mar 20, 2013 #7

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    Would I be correct in saying this sequence has a lower bound of 0 but no upper bound?
     
  9. Mar 21, 2013 #8

    LCKurtz

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    Is ##a_n>0## for all n? What do you think? Is that the greatest lower bound?
    Is ##a_n > n##? Again, what do you think?
     
  10. Mar 21, 2013 #9

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    Yes I believe that 0 is the greatest lower bound. And after plotting a couple of points on a graph of this sequence it appears this is no least upper bound.
     
  11. Mar 21, 2013 #10

    LCKurtz

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    The problem with "beliefs" is that they are sometimes wrong. And saying "it appears" is a long way from a proof. If this is a homework problem, you have a ways to go before handing it in.
     
  12. Mar 21, 2013 #11

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    Well because there is not an m that is ≥ a sub n this tells me there is no least upper bound. I know the greatest lower bound is 0 because the sequence never goes below 0 and therefore there is an M that is ≤ to all a sub n. according to my class notes this is as much evidence as I can provide. That we have learned thus far.
     
  13. Mar 21, 2013 #12

    LCKurtz

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    That argument shows 0 is a lower bound, but not that it is the greatest lower bound.
     
  14. Mar 21, 2013 #13

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    Do you have a quick explanation as to what makes a bound the least or greatest? Not seeing how 0 isn't the greatest lower bound when the sequence cannot go below that.
     
    Last edited: Mar 21, 2013
  15. Mar 21, 2013 #14

    LCKurtz

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    What is the definition for x to be the greatest lower bound of a set S of numbers?
     
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