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Thread: 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 $\displaystyle p$-series $\displaystyle \sum_{n=2}^\infty \frac {1}{n^p}$ converges if $\displaystyle p>1$ and diverges otherwise.

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

    The answer is $\displaystyle p=1$ (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 $\displaystyle \frac 1{n (\ln n)^p}$ to $\displaystyle \frac {1}{n^p}$. If $\displaystyle \frac 1{(n \ln n)^p} < \frac {1}{n^p}$, then the former converges, too. For $\displaystyle p>1$, and since $\displaystyle n>1$, then $\displaystyle (\ln n)^p>1 $and thus $\displaystyle \frac 1{(n \ln n)^p} < \frac {1}{n^p}$. How am I doing so far?

    But what about when $\displaystyle p<1$? Maybe for some values close to, but less than 1, $\displaystyle \sum_{n=2}^\infty \frac 1{n (\ln n)^p}$ still converges. How do I find out?
    Last edited by MSUMathStdnt; Dec 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 $\displaystyle p$-series $\displaystyle \sum_{n=1}^\infty \frac {1}{n^p}$ converges if $\displaystyle p>1$ and diverges otherwise.

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

    The answer is $\displaystyle p=1$ (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.

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

    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 $\displaystyle 2$, not $\displaystyle 1$. If you try to take the $\displaystyle \ln(1)$ in the denominator... well, you know.

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

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

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

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

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