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Math Help - prove this result please

  1. #1
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    prove this result please

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  2. #2
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    result

    i know this result, alpha <= 1, it is then divergent...but how to prove that?
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  3. #3
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    When \alpha > 1, use the integral test.
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  4. #4
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    cheers

    cheers mate!!!
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by szpengchao View Post
    I

    \sum\frac{1}{n\ln(n)}

    \int_3^{\infty}\frac{dn}{n\ln(n)}=\ln(\ln(n))\bigg  |_3^{\infty}=\infty

    divergent

    ii

    \sum_{n=3}^{\infty}\frac{1}{n\ln(n)(\ln\ln(n)))^2}

    \int\frac{dn}{n\ln(n)(\ln\ln(n)))^2}=\frac{-1}{\ln\ln(n))}\bigg|_3^{\infty}=\frac{1}{\ln(\ln(3  ))}

    convergent

    iii

    \sum_{n=3}^{\infty}\frac{1}{n^{1+\frac{1}{n}}\ln(n  )}

    Limit comparison test with \frac{1}{n\ln(n)}

    \lim_{n\to\infty}\frac{\frac{1}{n\ln(n)}}{\frac{1}  {n^{1+\frac{1}{n}}\ln(n)}}=\lim_{n\to\infty}n^{\fr  ac{1}{n}}\cdot\lim_{n\to\infty}\frac{n\ln(n)}{n\ln  (n)}

    Side note L=\lim_{n\to\infty}n^{\frac{1}{n}}\Rightarrow{ln(L  )=\lim_{n\to\infty}\frac{\ln(n)}{n}=0}

    so \ln(L)=0\Rightarrow{L=e^0=1}

    so \lim_{n\to\infty}n^{\frac{1}{n}}\cdot\lim_{n\to\in  fty}\frac{n\ln(n)}{n\ln(n)}=1\cdot{1}=1

    (EDIT: I should rephrase so you dont misinterpret) Since the limit converges to a finite value and the comparison series diverges, this series diverges
    Last edited by ThePerfectHacker; May 21st 2008 at 06:13 PM.
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