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Thread: Series Question

  1. #1
    Member RedBarchetta's Avatar
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    Series Question

    $\displaystyle
    \sum\limits_{n = 1}^\infty {\frac{{5\ln n}}
    {{\sqrt {n^3 } }}} $

    Is there any easy way to find the convergence/divergence of this series?

    So far I tried using the integral test and came up with a convergence at 20.
    This involves integration by substitution. Is there any easier method?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by RedBarchetta View Post
    Is there any easier method?
    Maybe : $\displaystyle \sqrt{n^3}=n^\frac32=n^\frac54n^\frac14$ hence

    $\displaystyle
    5\sum_{n = 1}^\infty\frac{\ln n}{\sqrt {n^3 }}=5\sum_{n = 1}^\infty\frac{1}{n^\frac54}\times\frac{\ln n}{n^\frac14}$.

    As $\displaystyle \frac{\ln n}{n^\frac14}\xrightarrow[n\to\infty]{}0$ one has, for $\displaystyle n$ sufficiently large, $\displaystyle \left|\frac{\ln n}{n^\frac14}\right|\leq 1$ hence $\displaystyle \left|\frac{1}{n^\frac54}\times\frac{\ln n}{n^\frac14}\right|\leq\left|\frac{1}{n^\frac54}\ right|\times 1 = \left|\frac{1}{n^\frac54}\right|$ that is to say $\displaystyle \left|\frac{\ln n}{\sqrt {n^3 }}\right|\leq\left|\frac{1}{n^\frac54}\right|$. As $\displaystyle \sum\frac{1}{n^\frac54}$ is an absolutely convergent series, the series $\displaystyle 5\sum_{n = 1}^\infty\frac{\ln n}{\sqrt {n^3 }}$ converges too. (comparison test)
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  3. #3
    Member RedBarchetta's Avatar
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    Thanks.
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