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Math Help - Analize convergence of an infinite series

  1. #16
    MHF Contributor kalagota's Avatar
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    not that fast..

    look what you have posted before my first post here..
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  2. #17
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Moo View Post
    Why ?

    \sum (a_n+b_n)=a_1+b_1+a_2+b_2+\dots

    \sum a_n+\sum b_n=a_1+a_2+\dots+b_1+b_2+\dots

    rearranging the terms won't change the convergence(...) or divergence, so :

    \sum a_n+\sum b_n=a_1+b_1+a_2+b_2+\dots=\sum (a_n+b_n)

    The most important thing here, imo, is that the domain of n for the two sums is the same, isn't it ?
    (...) if \sum (a_n+b_n) is absolutely convergent..
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  3. #18
    MHF Contributor kalagota's Avatar
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    example again:

    a_n = \frac{1}{n^2}
    b_n = \frac{1}{n}

    \sum \frac{1}{n^2} is convergent.
    \sum \frac{1}{n} is divergent.

    according to this:
    Quote Originally Posted by Moo View Post
    The sum of a divergent series and a convergent series is indeed divergent
    \sum \frac{1}{n^2} - \sum \frac{1}{n} must be divergent..

    how about \sum \frac{1-n}{n^2} = \sum \left(\frac{1}{n^2} - \frac{1}{n}\right)? will that be divergent? check it..
    Last edited by kalagota; July 21st 2008 at 02:30 AM.
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  4. #19
    MHF Contributor kalagota's Avatar
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    what i mean there is that your statement in red is only true for absolutely convergent series.. and not generally..
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  5. #20
    Moo
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    Actually, I'm completely confused

    I'll think about it this afternoon.
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  6. #21
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    Cool, now I'm confused too.
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  7. #22
    MHF Contributor kalagota's Avatar
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    no need to be confused.
    what i am saying is that the equality
    \sum (a_b + b_n) = \sum a_n + \sum b_n
    is always true if both \sum a_n and \sum b_n are convergent.
    and it is not true if at least 1 is divergent..
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  8. #23
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    Oh yeah, that's right. Thanks
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  9. #24
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Actually, I'm completely confused

    I'll think about it this afternoon.
    For you to be able to manipulate a series in the manner you are referring to it must be Cesaro Summable, and one of those conditions is that the series is convergent.

    An example where rearanging is not permissable is the following

    x=1-1+1-1\cdots

    \Rightarrow{x=1-(1-1+\cdots)}

    \Rightarrow{x=1-x}

    \Rightarrow{x=\frac{1}{2}}

    So

    \sum_{n=0}^{\infty}(-1)^n=\frac{1}{2}
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  10. #25
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    Quote Originally Posted by Moo View Post
    \sum (a_n+b_n)=a_1+b_1+a_2+b_2+\dots

    \sum a_n+\sum b_n=a_1+a_2+\dots+b_1+b_2+\dots

    rearranging the terms won't change the convergence or divergence, so :

    \sum a_n+\sum b_n=a_1+b_1+a_2+b_2+\dots=\sum (a_n+b_n)

    The most important thing here, imo, is that the domain of n for the two sums is the same, isn't it ?
    If a_n and b_n are convergent then \Sigma_{n=1}^{\infty}(a_n+b_n) = \Sigma_{n=1}^{\infty} a_n + \Sigma_{n=1}^{\infty}b_n (no absolute convergence is used here).

    If a_n or b_n are divergent then it does not even make sense to write \Sigma_{n=1}^{\infty}a_n + \Sigma_{n=1}^{\infty}b_n. Because what is that even supposed to mean?

    For you to be able to manipulate a series in the manner you are referring to it must be Cesaro Summable, and one of those conditions is that the series is convergent.

    An example where rearanging is not permissable is the following
    There is no rearrangment going on over here (in what Moo did). Just term by term addition.

    Let a_n and b_n be convergent series. Then s_n = a_1+...+a_n and t_n = b_1+...+b_n are convergent sequences. Now if s_n and t_n are convergent it means s_n+t_n is a convergent. Furthermore, \lim ~ (s_n+t_n) = \lim ~ s_n + \lim ~ t_n. By definition, \lim ~ s_n = \Sigma_{n=1}^{\infty}a_n and \lim ~ t_n = \Sigma_{n=1}^{\infty}b_n. And \lim ~ (s_n+t_n) = \Sigma_{n=1}^{\infty} (a_n+b_n). Thus, \Sigma_{n=1}^{\infty}(a_n+b_n) = \Sigma_{n=1}^{\infty} a_n + \Sigma_{n=1}^{\infty}b_n.
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  11. #26
    Moo
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    Ok, I've talked to someone. Things are clearer

    - \sum (a_n+b_n)=\sum a_n+\sum b_n if both are convergent infinite series, or if both are finite series.
    - the sentence for "sum of a divergent+a convergent is divergent" is true according to this person. Can anyone confirm or infirm ?
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  12. #27
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    Quote Originally Posted by Moo View Post
    - the sentence for "sum of a divergent+a convergent is divergent" is true according to this person. Can anyone confirm or infirm ?
    That is easy. If you combine something that is finite and infinite you get something that is infinite. So what is the problem?




    Just joking! That is how they "prove" in physics courses. Say \Sigma a_n is convergent and \Sigma b_n is divergent. And assume that \Sigma (a_n+b_n) is convergent. Then it would mean \Sigma (a_n+b_n) - a_n = \Sigma b_n is convergent. A contradiction!
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  13. #28
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Just joking! That is how they "prove" in physics courses. Say \Sigma a_n is convergent and \Sigma b_n is divergent. And assume that \Sigma (a_n+b_n) is convergent. Then it would mean \Sigma (a_n+b_n) - a_n = \Sigma b_n is convergent. A contradiction!
    EDIT:i misread your proof last night.. so probably, i should erase my previous comment..
    Last edited by kalagota; July 23rd 2008 at 04:50 AM.
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