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Math Help - Test for convergence...

  1. #1
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    Test for convergence...

    Test the series for convergence or divergence.

    The sum figure from K=1 to infinity 2 / ((k^2)*(1+k^2))^(1/3)
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  2. #2
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    \frac{1}{\sqrt[3]{k^{2}\left( 1+k^{2} \right)}}<\frac{1}{k^{4/3}}, so what we conclude?
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  3. #3
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    Convergent. Thanks! Why did you use 4/3?
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  4. #4
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    from what i remember if it`s goes to zero faster then 1/x it is convergencing
    and it does

    (don`t have the proof with me)
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  5. #5
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    ah, it's quite simple.

    since it's for each k\ge1, then it's obvious that \frac{1}{k^{2}\left( 1+k^{2} \right)}<\frac{1}{k^{2}\cdot k^{2}}=\frac{1}{k^{4}}, and you'll get the inequality.
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