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Math Help - Analysis (Series Convergence)

  1. #1
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    Analysis (Series Convergence)

    Show that the SERIES log k / (k^p), p>1 converges.
    thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by thahachaina View Post
    Show that the SERIES log k / (k^p), p>1 converges.
    thanks.
    Let \sum{a_n} be an infinite series with a_{n+1}\leqslant{a_n} and a_n>0. Then \sum{a_n} converges iff \sum{2^na_{2^n}} converges. So since \forall{k}\in(2,\infty)~\frac{\log(k)}{k^p}>1 and monotonically decreasing we may apply the above test. So

    \sum2^n\frac{\log(2^n)}{\left(2^n\right)^p}=\sum\f  rac{n\log(2)}{2^{n(p-1)}}. Now applying the Root test gives

    \begin{aligned}\limsup\sqrt[n]{\left|\frac{n\log(2)}{2^{n(p-1)}}\right|}&=\limsup\frac{\sqrt[n]{n\log(2)}}{2^{p-1}}\\<br />
&=\frac{1}{2^{p-1}}\end{aligned}

    So now for a series to converge we must have that \limsup\sqrt[n]{|a_n|}<1

    So solving this for p gives p>1
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by thahachaina View Post
    Hey guys,

    Does the SERIES, log k/ k^p, where p>1, converge or diverge???

    Thanks
    It converges because \log(k) increases more slowly than k^{(p-1)/2} and so eventually:

     <br />
\frac{\log(k)}{k^p}<\frac{1}{k^{\frac{p+1}{2}}}=\f  rac{1}{k^q}<br />

    and q=(p+1)/2>1.

    CB
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