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Math Help - Series(convergence divergence)

  1. #1
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    Series(convergence divergence)

    The asumptions are: an>0 (an is a sequence, where n is a subscript) and Sum(an) diverges

    What can be said about Sum(an/(1+n*an))

    I believe that this series can diverge and converge.

    I found that if we let an=(1/n), which satisfies our assumptions, then the Sum(an/(1+n*an))=Sum(an/2)=(1/2)Sum(an) which diverges.

    I cannot find an example of a diverding an series for which the Sum converges.
    Last edited by forget_f1; November 8th 2005 at 08:14 AM. Reason: Clarification
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  2. #2
    Grand Panjandrum
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    It seems to me that if a_n = 1/n then

    <br />
\sum_1^\infty\ \frac{a_n}{1+n\cdot a_n}\ =\ \sum_1^\infty\ \frac{1/n}{1+1}\ =\ \sum_1^\infty\ \frac{1}{2n}<br />

    which diverges, or have I misunderstood something?

    RonL
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  3. #3
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    That is correct for (1/n), but I cannot find an example for convergence.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by forget_f1
    The asumptions are: an>0 (an is a sequence, where n is a subscript) and Sum(an) diverges

    What can be said about Sum(an/(1+n*an))

    I believe that this series can diverge and converge.

    I found that if we let an=(1/n), which satisfies our assumptions, then the Sum(an/(1+n*an))=Sum(an/2)=(1/2)Sum(an) which diverges.

    I cannot find an example of a diverding an series for which the Sum converges.
    Why do you believe that with these assumptions about  a_n that

     \sum_1^\infty\ \frac{a_n}{1+n\cdot a_n}

    can converge?
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  5. #5
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    By the way, thank you for your interest and time. The reason why I believe it may converge is because the series in question is half the harmonic
    mean of the series a_n and the series 1/n, both of which are divergent.
    Any mean (including the harmonic mean) should be expected to be
    between the two numbers, which means that the series in question (or
    rather twice the series in question) should be bigger than either a_n
    or 1/n, whichever is smaller.

    The trouble is that you can't tell whether a_n will be bigger or 1/n,
    and it might differ for different n. For example, if it was always
    a_n, or if it was always 1/n, then it would be easy to prove that the
    sum will always diverge if an does. But what if it changes? Maybe a_n
    is smaller only sometimes. In fact, maybe a_n is very small (say
    1/n^2) for most values of n, but rather large (say 1) for occasional
    rare values of n (say perfect squares). Then a_n still diverges, but
    the new series is very small for most values, and not quite as large
    as a_n for the rare values, perhaps still small enough to converge.
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  6. #6
    Grand Panjandrum
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    I hope this provides the example you seek.

    Consider the following sequence

    a_n\ =\ \left\{\begin{array}{cc}1,&\mbox{ if } n = k! \mbox{ for k in N}\\0, & \mbox{ otherwise }\end{array}\right

    then the series: <br />
\sum_1^\infty\ a_n<br />
diverges.

    Also:

    <br />
\sum_1^\infty\ \frac{a_n}{1+n\cdot a_n} \\ = \sum_1^\infty\ \frac{1}{1+n!}<br />

    which does converge.

    Now all that remains to provide your example is to replace the
    zero terms in the equence a_n with terms from
    a convergent series of positive terms.

    RonL
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  7. #7
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    Thank you for your help, time and effort.
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