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Math Help - For what Values to the series converge

  1. #1
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    For what Values to the series converge

    I am struggling with this one. For what values for p does the series converge?

    \sum_{n=1}^\infty  ln(n)/n^p
    Last edited by zifnib; November 23rd 2008 at 08:12 PM. Reason: get sum in there
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by zifnib View Post
    I am struggling with this one. For what values for p does the series converge?

    \sum_{n=1}^\infty  ln(n)/n^p
    Hint: \exists{N}\backepsilon\forall{n\geqslant{N}}~\ln(n  )\leqslant{n^{\varphi>0}}

    Or use the Ratio test
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  3. #3
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    I have tried the ratio test but I keep getting 1(inconclusive)

    I am not sure by what you typed but it looks like you said its true for any p > 0 but when p=1 then  ln(n)/n and that is divergent. But when p = 2 or anything greater it looks like it is convergent. I need a way to show this. I think that if I can show that when p =2 that that is convergent than by comparison any p > 2 is convergent but that may be incorrect as well???
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Your right about the Ratio test sorry, but why not try the integral test

    \int_1^{\infty}\frac{\ln(n)}{n^p}dn=\lim_{n\to\inf  ty}\bigg[\frac{1}{1-p}n\ln(n)-\frac{1}{1-2p+p^2}n\bigg]n^{-p}

    Now find for what values that is convergent.
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  5. #5
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    I tried that on my TI-89 but I was not sure to to tell which values for p would work that way? Would I just need to find where the limit is a number other than 0 or infinity? Something seams incorrect to me kinda.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by zifnib View Post
    I tried that on my TI-89 but I was not sure to to tell which values for p would work that way? Would I just need to find where the limit is a number other than 0 or infinity? Something seams incorrect to me kinda.
    Don't use your TI-89! Use your brain! Distribute n^{-p} and notice what happens.

    Here is one final way you could prove it, since your series is decreasing and positive after some value of N (use the inequality Ifirst gaave you to prove that) then your series covnerges iff \sum_{n=1}^{\infty}2^n\cdot\frac{\ln\left(2^n\righ  t)}{2^{np}}=\sum_{n=1}^{\infty}\frac{\ln(2)n}{2^{(  p-1)n}}

    Now think what values of p does this serise converges...this time its pretty apparent.
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