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Math Help - Another Question for a P-Series Problem

  1. #1
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    Another Question for a P-Series Problem

    Determine all values of p for which the series is convergent, and express answer in interval notation.

    \sum_{n=2}^{\infty} (-1)^{n-1}  \frac{ln(n)}{2n}^{p}

    How would I find the values of p that makes this series a convergent one. It appears to be an alternating series, so would I just focus on \frac{ln(n)}{2n}^{p} ?
    I am just not quite sure how to go about this problem. Could I used the ratio test or anything else?

    Any feedback and help appreciated, thanks.
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    Lightbulb Re: Another Question for a P-Series Problem

    Quote Originally Posted by Beevo View Post
    Determine all values of p for which the series is convergent, and express answer in interval notation.
    \sum_{n=2}^{\infty} (-1)^{n-1}  \frac{ln(n)}{2n}^{p}
    For p=1000 the series converges to 4.312152513533860\times 10^{554}

    It appears that the series converges for all p<\infty

    Attached Thumbnails Attached Thumbnails Another Question for a P-Series Problem-p-series.png  
    Last edited by MaxJasper; September 10th 2012 at 12:24 PM.
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    Re: Another Question for a P-Series Problem

    \frac{\ln{n}^p}{2n} = \frac{p}{2} \cdot \frac{\ln{n}}{n}

    since \lim_{n \to \infty} \frac{\ln{n}}{n} = 0 , and the series is alternating ...

    \sum_{n=2}^{\infty} (-1)^{n-1}\frac{\ln{n}^p}{2n} = \frac{p}{2} \sum_{n=2}^{\infty} (-1)^{n-1} \frac{\ln{n}}{n}

    ... is conditionally convergent for all fixed values of p.
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    Re: Another Question for a P-Series Problem

    Quote Originally Posted by skeeter View Post
    \frac{\ln{n}^p}{2n} = \frac{p}{2} \cdot \frac{\ln{n}}{n}

    since \lim_{n \to \infty} \frac{\ln{n}}{n} = 0 , and the series is alternating ...

    \sum_{n=2}^{\infty} (-1)^{n-1}\frac{\ln{n}^p}{2n} = \frac{p}{2} \sum_{n=2}^{\infty} (-1)^{n-1} \frac{\ln{n}}{n}

    ... is conditionally convergent for all fixed values of p.
    I came to this same conclusion. Since it is an alternating series, I took the limit of bn and found that it was converging.

    However, there is also the rule that the series has to be decreasing for it to be convergent in a alternating series. So, would the value of p affect the rule of decrease? That is what I am stuck on. If not, then clearly the interval notation for the value of p that makes this series converge is [o to infinity).
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    Re: Another Question for a P-Series Problem

    note that p/2 is a constant that can be factored out of the series terms ...

    k \sum b_n where k is a constant is convergent if \sum b_n is convergent.
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    Re: Another Question for a P-Series Problem

    Quote Originally Posted by skeeter View Post
    note that p/2 is a constant that can be factored out of the series terms ...

    k \sum b_n where k is a constant is convergent if \sum b_n is convergent.
    Got it!!! Thanks man.

    I did not realize that the p could be factored out to form a constant. Makes sense now.
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    Question Re: Another Question for a P-Series Problem

    I suspect the correct term in the series might be: [\ln (n)]^p
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    Re: Another Question for a P-Series Problem

    Quote Originally Posted by MaxJasper View Post
    I suspect the correct term in the series might be: [\ln (n)]^p
    Yeah, the entire natural log function is raised to the p. Have I been solving it incorrectly?
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    Re: Another Question for a P-Series Problem

    Others assumed ln[n^p]
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    Re: Another Question for a P-Series Problem

    Oh really? That's a problem. Then I have been solving it incorrectly. Well it is still an alternating series, I just need to figure out what values of p make this series convergent. Any ideas?
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