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Limit of this sequence.

  1. Aug 22, 2011 #1
    I have a sequence {xn} defined by

    xn = 1/n[1 + 1/2 + 1/3 + ... + 1/n]
    for all natural numbers n.

    I want to show that this sequence converges to 0, i.e. given any positive real number 'r', I want to show that there exists a natural number k such that xk < r. (The sequence is monotonically decreasing.)

    Help please.
     
  2. jcsd
  3. Aug 22, 2011 #2

    Dick

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    Find an upper bound for [1 + 1/2 + 1/3 + ... + 1/n] using an integral to estimate it. That would be a good start.
     
    Last edited: Aug 22, 2011
  4. Aug 22, 2011 #3
    Alternatively, you could prove the general case:
    [tex]
    a_n \rightarrow a \Rightarrow \frac{1}{n} \displaystyle \sum_{k=1}^{n} a_k \rightarrow a
    [/tex]
     
  5. Aug 22, 2011 #4

    hunt_mat

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    How about an application of the sandwich theorem.
     
  6. Aug 23, 2011 #5
    Hi!
    [1 + 1/2 + ... + 1/n] is always less than [1+ln(n)] ?
    implying that my sequence is less that [1/n + (1/n)*ln(n)] which I believe converges to 0.
     
  7. Aug 23, 2011 #6

    Dick

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    All true. But how did you show the sum is less than 1+ln(n)? And 'believing' [1/n + (1/n)*ln(n)] converges to zero doesn't prove it does. How would you show that?
     
  8. Aug 24, 2011 #7
    Isn't it something like the integral test? It was kinda obvious from the diagram.
    And lim [x^x] -> 1 as x->0 follows from L'Hopital's rule. (The log of the limit is 0)
     
  9. Aug 24, 2011 #8
    you could use the divergence test
     
  10. Aug 24, 2011 #9

    Dick

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    Yes, it is l'Hopital, but I don't see what x^x has to do with it. And yes, the integral part is sort of obvious from a lower sum estimate of 1/x. If you don't want to actually spell out the details, that's fine with me.
     
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