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Math Help - Limit comparison test

  1. #1
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    Limit comparison test

    How do you do these using the limit comparison test?

    • \sum_{n=1}^\infty \frac{1}{n(n^2+1)}


    \sum_{n=1}^\infty \frac{1}{n^2} p=2 > 1 converges

    \lim_{n\rightarrow\infty} \frac {1}{n(n^2+1)} * \frac{n^2}{1} <br />
= \lim_{n\rightarrow\infty} \frac{n^2}{n^3+n} ??



    • \sum_{n=1}^\infty \frac{3}{n(n+3)}


    \sum_{n=1}^\infty \frac{1}{n} p=1 diverges

    \lim_{n\rightarrow\infty}...=\\ \lim_{n\rightarrow\infty} \frac {3n}{n^2+3n} ??

    Thanks
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  2. #2
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    Quote Originally Posted by JJ007 View Post
    How do you do these using the limit comparison test?

    • \sum_{n=1}^\infty \frac{1}{n(n^2+1)}
    \sum_{n=1}^\infty \frac{1}{n^2} p=2 > 1 converges

    \lim_{n\rightarrow\infty} \frac {1}{n(n^2+1)} * \frac{n^2}{1} <br />
= \lim_{n\rightarrow\infty} \frac{n^2}{n^3+n} ?? = ZERO ---> The series converges.



    • \sum_{n=1}^\infty \frac{3}{n(n+3)}
    \sum_{n=1}^\infty \frac{1}{n} p=1 diverges

    \lim_{n\rightarrow\infty}...=\\ \lim_{n\rightarrow\infty} \frac {3n}{n^2+3n} ?? = ZERO ---> Test failed. Make a limit comparison test with the convergent series: \sum_{n \geq 1} \frac{1}{n^2}

    Thanks
    ..
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