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Math Help - [SOLVED] Use ratio test to determine if series converges or diverges

  1. #1
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    [SOLVED] Use ratio test to determine if series converges or diverges

    \sum_{n=1}^{\infty}\ \frac{k!}{k^3}
    I'm stuck about half way through
    this is what ive done so far
    lim_{k->\infty}\frac{(k+1)!}{(k+1)^3}*\frac{k^3}{k!}
    I know that getting rid of the factorials gives me k+1 in the numerator but im not sure how to simplify the k^3and (k+1)^3
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is \frac{(k+1)!}{(k+1)^{3}}\cdot \frac{k^{3}}{k!} = (k+1)\cdot (1-\frac{1}{k})^{3} so that...

    Kind regards

    \chi \sigma
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  3. #3
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    Hello, vinson24!

    \sum_{n=1}^{\infty}\ \frac{k!}{k^3}

    I'm stuck about half way through

    This is what ive done so far: . \lim_{k->\infty}\frac{(k+1)!}{(k+1)^3}\cdot\frac{k^3}{k!  }

    I know that getting rid of the factorials gives me k+1 in the numerator
    but im not sure how to simplify the k^3 and (k+1)^3

    We have: . \frac{(k+1)!}{k!}\cdot\frac{k^3}{(k+1)^3} \;=\;\frac{k+1}{1}\cdot\left(\frac{k}{k+1}\right)^  3

    In the parentheses, divide top and bottom by k\!:\quad \frac{k+1}{1}\cdot\left(\frac{1}{1+\frac{1}{k}}\ri  ght)^3

    Got it?

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  4. #4
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    yes diverges thanks
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