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Converges or diverges

  1. Apr 19, 2005 #1
    Determine whether the sequence [tex]a_n = \frac{1^2}{n^3} + \frac{2^2}{n^3} + ... + \frac{n^2}{n^3} [/tex] converges or diverges. If it converges, find the limit.


    wouldnt it converge to 0?
     
  2. jcsd
  3. Apr 19, 2005 #2

    Hurkyl

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    Maybe, maybe not. Why do you think it will go to 0? Any ideas on how to go about trying to prove it?
     
  4. Apr 19, 2005 #3

    James R

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    That seems like a difficult problem, since you have a sequence of series.

    The first few terms in the sequence would be:

    [tex]\left{1, \frac{1^2}{2^3} + \frac{2^2}{2^3}, \frac{1^2}{3^3} + \frac{2^2}{3^3} + \frac{3^2}{3^3}, \frac{1^2}{4^3} + \frac{2^2}{4^3} + \frac{3^2}{4^3} + \frac{4^2}{4^3},...\right}[/tex]
     
  5. Apr 19, 2005 #4

    quasar987

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    The nth term is increasing and bounded by one

    [tex]a_n \leq \frac{n^2}{n^3} + \frac{n^2}{n^3} + ... + \frac{n^2}{n^3} = n\frac{n^2}{n^3} = 1[/tex]
     
  6. Apr 19, 2005 #5

    quasar987

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    Actually, the nth term can be rewritten as

    [tex] a_n = \frac{1}{n^3}\sum_{i=1}^{n}i^2 [/tex]

    Do you recognize this sum?
     
  7. Apr 19, 2005 #6

    quasar987

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    The reason why the value of convergence of this sequence is not simply obtained by doing

    [tex]\lim_{n\rightarrow \infty} \frac{1^2}{n^3} + \frac{2^2}{n^3} + ... + \frac{n^2}{n^3} = \lim_{n\rightarrow \infty} \frac{1^2}{n^3} + \lim_{n\rightarrow \infty} \frac{2^2}{n^3} + ... + \lim_{n\rightarrow \infty} \frac{n^2}{n^3} = 0+0+...+0[/tex]

    is that the NUMBER OF TERMS in the sum augment to infinity as well! So you can't be sure what the sum's gonna be: even though all members of the sum go to zero, there is an infinity of them.
     
  8. Apr 20, 2005 #7
    so it converges, but no answer?
     
  9. Apr 20, 2005 #8
    just a thought, wouldnt the limit comparison test work on this if you choose your b_n to be 1/n, and use the p-series to say b_n diverges?
     
  10. Apr 21, 2005 #9

    Hurkyl

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    Pro: we're here to help, not give answers. You should have enough information to figure it out yourself.

    mug: what are you comparing to 1/n?
     
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