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Math Help - contradictions in convergence and divergence rules. need clarification

  1. #1
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    contradictions in convergence and divergence rules. need clarification

    hello, so i'm reviewing the chapter on absolute and conditional divergence, and this chapter has me completely baffled.

    are p-series not always applicable?

    because, for example,

    infinite
    sigma [(-1)^n]/(n)^1/2
    n=1

    ace of n = 1/n^(1/2)

    converges, even though p=1/2 < 1.

    but at the same time,

    infinite
    sigma 1/n^(1/2)
    n=1

    is divergent because the p-series somehow applies to this one and not that one.

    are p-series only applicable for non-alternating series? when does the leibniz rule override the p-series rule?

    also, for

    infinite
    sigma [(-1)^n]/[(n^2)+1]^1/2
    n=1

    converges. does that mean


    infinite
    sigma 1/[(n^2)+1]^1/2
    n=1

    converges?

    andy why is ace of n called a sequence that converges to zero when it is NOT alternating and is therefore supposed to diverge?

    i'm very confused.
    Last edited by wyhwang7; March 15th 2009 at 04:01 AM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by wyhwang7 View Post
    hello, so i'm reviewing the chapter on absolute and conditional divergence, and this chapter has me completely baffled.

    are p-series not always applicable?

    because, for example,

    infinite
    sigma [(-1)^n]/(n)^1/2
    n=1

    ace of n = 1/n^(1/2)

    converges, even though p=1/2 < 1.

    but at the same time,

    infinite
    sigma 1/n^(1/2)
    n=1

    is divergent because the p-series somehow applies to this one and not that one.

    are p-series only applicable for non-alternating series? when does the leibniz rule override the p-series rule?
    Yes. If you check your text you should see that a "p" test is that all the terms be postive. If by the "Leigniz" rule, you mean that " \sum(-1)^na_n converges whenever a_n goes to 0", it doesn't have to "override" the p-test because it only applies to alternating series while the p-test only applies to positive series (if all terms are negative you can factor out -1 so more generally it applies to series where the terms are all of the same sign).

    also, for

    infinite
    sigma [(-1)^n]/[(n^2)+1]^1/2
    n=1

    converges. does that mean


    infinite
    sigma 1/[(n^2)+1]^1/2
    n=1

    converges?
    No it doesn't.

    andy why is ace of n called a sequence that converges to zero when it is NOT alternating and is therefore supposed to diverge?

    i'm very confused.
    A sequence and a series are very different things. Nor is it true that any series that is not alternating diverges.
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