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Math Help - comparison test sum exists

  1. #1
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    comparison test sum exists

    I know that:

    this sequnce \sum _{k=1}^{\infty }{k}^{-1} doesnot exist

    but this sequence \sum _{k=1}^{\infty }-{k}^{-2}does exist

    How can i prove this sequence \sum _{k=1}^{\infty }{{k}^{-1}-{k}^{-2}} does not exists
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  2. #2
    Super Member Showcase_22's Avatar
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    <br /> <br />
\sum _{k=1}^{\infty }\left( {{k}^{-1}-{k}^{-2}} \right)=\sum_{k=1}^{\infty} \frac{1}{k}-\sum_{k=1}^{\infty} \frac{1}{k^2}<br />

    We know that \sum_{k=1}^{\infty} \frac{1}{k} \rightarrow \infty and \sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}

    so the RHS approaches \infty meaning that the series does not converge.
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