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Convergence of a sequence

  1. Oct 15, 2004 #1

    quasar987

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    Apparently (according to my textbook), the sequence defined by

    [tex]\left\{\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right\}[/tex]

    converges towards 1/2, i.e. has 1/2 as a limit.

    How could that be?! It seems to me that as n approaches infinity, all the fractions fall to zero. What is it I'm missing?
     
  2. jcsd
  3. Oct 15, 2004 #2

    Hurkyl

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    The question is if they tend to zero faster than their number grow towards infinity.
     
  4. Oct 15, 2004 #3

    quasar987

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    :eek:

    Is there a way to find this analytically?
     
  5. Oct 15, 2004 #4

    arildno

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    Sure; you may write the partial sum as:
    [tex]\frac{1}{n^{2}}(1+2+++n-1)=\frac{1}{n^{2}}\frac{n(n-1)}{2}[/tex]
     
  6. Oct 15, 2004 #5

    quasar987

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    Gauss's sum again! Damn! You guys are smart, are you all doctors in mathematics or physics?
     
  7. Oct 15, 2004 #6
    Some of them are. :) Don't worry about it, I feel the same way you do all the time.
     
  8. Oct 15, 2004 #7

    quasar987

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  9. Oct 26, 2004 #8

    Tom Mattson

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    Hold on a second. How is it that the index appears in every term when you list out the series?

    Also, the above is a series, not a sequence.

    The limit of the sequence is zero.
    The limit of the sequence of partial sums is 1/2.
     
  10. Oct 26, 2004 #9

    shmoe

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    Tom, it's precisely the fact that the index "n" appears in each of the terms that makes this a sequence, and not a series, as it's given.

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

    It's the limit of [tex]a_n[/tex] he's after. Since each of the terms in the sum is dependant on n, you can't break it into a series as I suspect you are thinking of doing.


    You can of course think of any sequence as a series, by setting [tex]b_1=a_1, b_n=a_n-a_{n-1}[/tex], then [tex]a_n=\sum_{i=1}^{n}b_i[/tex], but that can be an awkward thing to do. In this case we'd find [tex]b_n=\frac{1}{2n(n+1)}[/tex], but I don't think that's what you were getting at?
     
    Last edited: Oct 26, 2004
  11. Oct 26, 2004 #10

    Tom Mattson

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    I really do know better than that...

    Do me a favor and just ignore me for the rest of the night....
     
    Last edited: Oct 26, 2004
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