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Math Help - [SOLVED] series

  1. #1
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    [SOLVED] series


    I'm trying to find the values of p for which the series converges (in interval notation). I used the integral test and came up with:




    if p<=-1 the series should be convergent, right?
    However, I entered (-inf, 0) as my answer and it came up incorrect. Could someone please help/correct me?
    It would be much appreciated! Thank you!
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  2. #2
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    Quote Originally Posted by yzobel View Post

    I'm trying to find the values of p for which the series converges (in interval notation). I used the integral test and came up with:




    if p<=-1 the series should be convergent, right?
    However, I entered (-inf, 0) as my answer and it came up incorrect.


    You said it: p\leq -1\Longrightarrow (-\infty,-1] should be the answer...

    Tonio

    Could someone please help/correct me?
    It would be much appreciated! Thank you!
    .
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  3. #3
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    I only have a limited number of answer choices though.
    (0, inf)
    (-inf, 0) <==what I thought it was
    (-inf, inf)
    [0, inf)
    p={0}
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  4. #4
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    Quote Originally Posted by yzobel View Post
    I only have a limited number of answer choices though.
    (0, inf)
    (-inf, 0) <==what I thought it was
    (-inf, inf)
    [0, inf)
    p={0}

    Well, now checking the problem in deep: you could use the integral test only for positive series, and this is not such, so what you did is to evaluate when the series converges absolutely...!

    Now check for which p's the series is a Leibnitz one (i.e., the series' sequence, without the sign, must be positive and monotone converging to zero)

    Tonio
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  5. #5
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    I don't think we've gone over Leibnitz series yet, but thank you anyways for all of your help!
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  6. #6
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    Leibniz's Theorem is the same thing as Alternating Series Test, it's just a different name.
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