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Math Help - What value of p does the following series converge

  1. #1
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    What value of p does the following series converge

    A Theorem say the p-series \sum_{n=2}^\infty \frac {1}{n^p} converges if p>1 and diverges otherwise.

    The question says "Find the positive values of p for which the series \sum_{n=2}^\infty \frac 1{n (\ln n)^p} converges.

    The answer is p=1 (from back of book). There is no explanation of how they got it and I have none either. How do you find that?

    I thought maybe I would compare \frac 1{n (\ln n)^p} to \frac {1}{n^p}. If \frac 1{(n \ln n)^p} < \frac {1}{n^p}, then the former converges, too. For p>1, and since n>1, then (\ln n)^p>1 and thus \frac 1{(n \ln n)^p} < \frac {1}{n^p}. How am I doing so far?

    But what about when p<1? Maybe for some values close to, but less than 1, \sum_{n=2}^\infty \frac 1{n (\ln n)^p} still converges. How do I find out?
    Last edited by MSUMathStdnt; December 19th 2012 at 05:23 PM.
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  2. #2
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    Re: What value of p does the following series converge

    Hey MSUMathStdnt.

    Have you tried using some other tests like the integral test?
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  3. #3
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    Re: What value of p does the following series converge

    Quote Originally Posted by MSUMathStdnt View Post
    A Theorem say the p-series \sum_{n=1}^\infty \frac {1}{n^p} converges if p>1 and diverges otherwise.

    The question says "Find the positive values of p for which the series \sum_{n=2}^\infty \frac 1{n (\ln n)^p} converges.

    The answer is p=1 (from back of book). There is no explanation of how they got it and I have none either. How do you find that?
    There is a true difficulty with your post.

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

    Look at the integral test.

    So what is your question now?
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  4. #4
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    Re: What value of p does the following series converge

    First a correction. The lower bounds was supposed to be 2, not 1. If you try to take the \ln(1) in the denominator... well, you know.

    OK, so here's the integral test (so far)

    \int_2^\infty \frac 1{n (\ln n)^p} dn = \frac 1{(1-p)(\ln n)^p} \bigg|_1^\infty

    The first term
    \lim_{n \to \infty} \frac 1{(1-p) (\ln n)^p} = 0 for all positive values of p \ne 1.

    The second term
    \lim_{n \to 2^+} \frac 1{(1-p) (\ln n)^p} = \frac 1{(1-p) (\ln 2)^p} for all p \ne 1.

    So the entire integral comes to -\frac 1{(1-p) (\ln 2)^p}, which means that the series converges when that fraction converges. Right so far? If so, now what?
    Last edited by MSUMathStdnt; December 19th 2012 at 05:30 PM.
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