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Math Help - convergence tests for series

  1. #1
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    Question convergence tests for series

    what would be an appropriate test (out of ratio, intergral, limit comparison) to decide if the series

    sum of(1 to infinity)[k^(1/2)/ (3*k^3+1)]

    converges or diverges?
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by Sniperpro View Post
    what would be an appropriate test (out of ratio, intergral, limit comparison) to decide if the series

    sum of(1 to infinity)[k^(1/2)/ (3*k^3+1)]

    converges or diverges?
    A (limit) comparison test would work best... certainly would not try the integral test.
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  3. #3
    MHF Contributor Amer's Avatar
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    you can use ratio test I prefer it

    lim_{k\rightarrow\infty}\frac{\sqrt{k+1}(3(k)^3)}{  ( \sqrt{k}(3(k+1)^3+1)}=lim_{k\rightarrow\infty}\fra  c{\sqrt{k+1}(3k^3)}{( \sqrt{k}(3k^3+3k^2+3k+1+1)}
    =lim_{k\rightarrow\infty}\frac{\frac{\sqrt{k+1}}{k  ^3}(3)}{\frac{\sqrt{k}}{k^3}(3+\frac{3}{k})+\frac{  3}{k^2}+\frac{2}{k^3})}
    =lim_{k\rightarrow\infty}\frac{\frac{\sqrt{k+1}}{k  ^3}(3)}{\frac{\sqrt{k}}{k^3}(3+\frac{3}{k})+\frac{  3}{k^2}+\frac{2}{k^3})}=1

    the.... test... failed ...try.. comparison... test
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  4. #4
    MHF Contributor Amer's Avatar
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    \sum{\frac{\sqrt{k}}{3k^3+1}}< \sum{\frac{k}{3k^3}}
    Last edited by Amer; May 27th 2009 at 01:12 PM.
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  5. #5
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    so would it be ok use k^(1/2)/k^3 as Bn in the limit comparison test?
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  6. #6
    MHF Contributor Amer's Avatar
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    No i think it will be the same take
    \frac{k}{3k^3}>\frac{\sqrt{k}}{3k^3+1}

    and..\frac{k}{3k^3}=\frac{1}{3k^2}

    \sum{\frac{1}{3k^2}} .converge .....by...integral
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