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Math Help - Need a help convergence!

  1. #1
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    Need a help convergence!

    My problem is Examine the series \sum_{n = 2}^\infty \frac{{n}^{lnn }}{{lnn}^{ n} } . I solved that:
    Applying the Cauchy's test we have:
    \sqrt[n]{\frac{{n}^{lnn }}{{lnn}^{ n} } } = \frac{{n}^{\frac{lnn}{ n} } }{lnn } and \lim_{n \to \infty} \frac{{n}^{\frac{lnn}{ n} } }{lnn } = 1 ( since let y = {n}^{\frac{lnn}{ n} we have lny = \frac{{lnn}^{2 } }{n }= \frac{2}{n }=0 while \lim_{n \to \infty}lnn= infinite => \lim_{n \to \infty}\frac{{n}^{lnn }}{{lnn}^{ n} }= 0 <1. Hence, the series converges. Is that right?
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  2. #2
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    Quote Originally Posted by ruanyueling View Post
    My problem is Examine the series \sum_{n = 2}^\infty \frac{{n}^{\ln n }}{{\ln n}^{ n}  } . I solved that:
    Applying the Cauchy's test we have:

    \sqrt[n]{\frac{{n}^{\ln n }}{{\ln n}^{ n}  }  } = \frac{{n}^{\frac{\ln n}{ n}  } }{\ln n }


    This is wrong: \sqrt[n]{\ln n^n}=\sqrt[n]{n\ln n}...


    and \lim_{n \to \infty} \frac{{n}^{\frac{\ln n}{ n}  } }{\ln n } = 1 ( since let y = {n}^{\frac{\ln n}{ n} we have

    \ln y = \frac{{\ln n}^{2 } }{n }= \frac{2}{n }=0, while \lim_{n \to \infty}\ln n= infinite => \lim_{n \to \infty}\frac{{n}^{\ln n }}{{\ln n}^{ n}  }= 0 <1.

    Hence, the series converges. Is that right?
    .


    Comparison seems to be way easier: \frac{n^{\ln n}}{n\ln n}\geq \frac{1}{\ln n} , and as the series of the rightmost expression diverges so does our series.

    Tonio
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  3. #3
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    ah, sorry. this is (lnn)^n, not lnn^n. ^^
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