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Math Help - Rookie question - Series 1/n

  1. #16
    iva
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    Thanks i realised i actually had this test all wrong... so can i for example take a series like

    a = 1/ root(3r)

    Can i use a p-series like 1/r^(1/3) ie 1/cuberoot(r) (call this b)

    which is divergent according to the p-series since p<1.

    And then I can deduce that since 0 < b < a and b is divergent, series a is divergent?
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  2. #17
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    Quote Originally Posted by iva View Post
    Thanks i realised i actually had this test all wrong... so can i for example take a series like

    a = 1/ root(3r)

    Can i use a p-series like 1/r^(1/3) ie 1/cuberoot(r) (call this b)

    which is divergent according to the p-series since p<1.

    And then I can deduce that since 0 < b < a and b is divergent, series a is divergent?
    Yes, that's correct. One way to remember the two aspects of the comparison test for series of positive terms is
    "(smaller than convergent) implies convergent,
    (larger then divergent) implies divergent."
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  3. #18
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    Quote Originally Posted by Opalg View Post
    Yes, that's correct. One way to remember the two aspects of the comparison test for series of positive terms is
    "(smaller than convergent) implies convergent,
    (larger then divergent) implies divergent."
    Alternatively:

    "Less than a number is a number" and "Greater than infinity is infinity..."
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  4. #19
    iva
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    Thanks, was just looking at my explanation again and i think its wrong, 1/cuberoot(r) gives a bigger number than 1/root(r). I guess i could use r^(-3/4) to make sure that its a power bigger than the orignal series (-1/2) but less than 1 right?
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  5. #20
    iva
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    Another question i had ( maybe the textbook is still going to explain..) is the scalar multiple of a divergent series, divergent , and convergent series, covergent? I think it is, because i guess a scalar multpile of it would be a subset of it.. am i right?
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  6. #21
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    Quote Originally Posted by iva View Post
    Another question i had ( maybe the textbook is still going to explain..) is the scalar multiple of a divergent series, divergent , and convergent series, covergent? I think it is
    That is right, multiplying each term of a series by a (nonzero!) constant has no effect on the convergence or divergence of the series.
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