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Math Help - Help with following infinite series: n/[(n^3)-1]

  1. #1
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    Help with following infinite series: n/[(n^3)-1]

    Hello,

    could anybody please help me with the following infinite series .. Σ[n/[(n^3)-1], n=2 .. infinite?
    i tried both the ratio test but it is equal to 1 so it is of no help, and the root test is useless, so what can i do to prove it converges? Does it converge?
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    Re: Help with following infinite series: n/[(n^3)-1]

    Quote Originally Posted by nappysnake;720594 Σ[n/[(n^3)-1
    , n=2 .. infinite? i tried both the ratio test but it is equal to 1 so it is of no help, and the root test is useless, so what can i do to prove it converges? Does it converge?
    Use the limit comparison test with \frac{1}{n^2}~.
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    Re: Help with following infinite series: n/[(n^3)-1]

    Quote Originally Posted by nappysnake View Post
    Hello,

    could anybody please help me with the following infinite series .. Σ[n/[(n^3)-1], n=2 .. infinite?
    i tried both the ratio test but it is equal to 1 so it is of no help, and the root test is useless, so what can i do to prove it converges? Does it converge?
    \sum_{n=2}^{\infty} \frac{n}{n^{3}-1}

     \frac{n}{n^{3}-1} = \frac{1}{n^{2} ( n-\frac{1}{n^{2}} ) } \le  \frac{1}{n^{2}} , n=2,3,4...

    \sum_{n=1}^{\infty} \frac{1}{n^{2}} is known to be convergent, so

    \sum_{n=22}^{\infty} \frac{n}{n^{3}-1} also converges.
    Last edited by fqqs; May 26th 2012 at 09:59 AM.
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    Exclamation Re: Help with following infinite series: n/[(n^3)-1]

    Yeah, i thought of the limit comparisson test as my last resource BUT  \frac{n}{n^{3}-1} is NOT equal  \frac{1}{n^{2} ( n-\frac{1}{n^{2}} ) } , n=2,3,4...

    BUT  \frac{n}{n^{3}-1} = \frac{n}{n^{2} ( n-\frac{1}{n^{2}} ) } = \frac{1}{n ( n-\frac{1}{n^{2}} ) }  \ge  \frac{1}{n^2} , n=2,3,4...
    therefor is useless for proving convergence because we require a fraction (series) for that matter which is greater. Am i right? Do i not see something?
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    Re: Help with following infinite series: n/[(n^3)-1]

    Quote Originally Posted by nappysnake View Post
    Yeah, i thought of the limit comparisson test as my last resource BUT  \frac{n}{n^{3}-1} is NOT equal  \frac{1}{n^{2} ( n-\frac{1}{n^{2}} ) } , n=2,3,4...
    \frac{{\frac{1}{{{n^2}}}}}{{\frac{n}{{{n^3} - 1}}}} = \frac{{{n^3} - 1}}{{{n^3}}} \to 1
    Thanks from nappysnake
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    Re: Help with following infinite series: n/[(n^3)-1]

    ok i just googled limit comparisson test and i've realized that the limit comparisson test is different than the direct comparisson test and i didn't know that. thank you very much for the help and clarification.
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