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Math Help - Series Question

  1. #1
    Member RedBarchetta's Avatar
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    Series Question

    <br />
\sum\limits_{n = 1}^\infty  {\frac{{5\ln n}}<br />
{{\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 : \sqrt{n^3}=n^\frac32=n^\frac54n^\frac14 hence

    <br />
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 \frac{\ln n}{n^\frac14}\xrightarrow[n\to\infty]{}0 one has, for n sufficiently large, \left|\frac{\ln n}{n^\frac14}\right|\leq 1 hence \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 \left|\frac{\ln n}{\sqrt {n^3 }}\right|\leq\left|\frac{1}{n^\frac54}\right|. As  \sum\frac{1}{n^\frac54} is an absolutely convergent series, the series 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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