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Thread: 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 $\displaystyle \sum{a_n}$ be an infinite series with $\displaystyle a_{n+1}\leqslant{a_n}$ and $\displaystyle a_n>0$. Then $\displaystyle \sum{a_n}$ converges iff $\displaystyle \sum{2^na_{2^n}}$ converges. So since $\displaystyle \forall{k}\in(2,\infty)~\frac{\log(k)}{k^p}>1$ and monotonically decreasing we may apply the above test. So

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

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

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

    So solving this for p gives $\displaystyle p>1$
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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 $\displaystyle \log(k)$ increases more slowly than $\displaystyle k^{(p-1)/2}$ and so eventually:

    $\displaystyle
    \frac{\log(k)}{k^p}<\frac{1}{k^{\frac{p+1}{2}}}=\f rac{1}{k^q}
    $

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

    CB
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