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Math Help - Series: comparison test

  1. #1
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    Series: comparison test

    Use the comparison test to investigate the convergence or divergence

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

    My solution:

    p-series \frac{1}{n^2} converges

    \frac{1}{n^2+1} < \frac{1}{n^2}


    then \sum_{n=1}^{\infty} \frac{1}{n^2+1} converges

    Is correct ?


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

    n! diverges in the ratio test, then I use the \lim_{n ->{\infty}} \frac{\frac{n!}{\sqrt{n(n+1)}}}{n!} = 0

    But is 0, how can I determine the convergence or divergence?
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  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
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    In problem 2:

    b_n = \frac{n!}{n} In the ratio test:

    \lim_{n \to \infty} \frac{(n+1)n!}{n+1}. \frac{n+1}{n!} = \infty Diverges

    Then, \frac{n!}{n} > \frac{n!}{\sqrt{n(n+1)}}

    \sum_{n=1}^{\infty} \frac{n!}{\sqrt{n(n+1)}} => Diverges

    It is correct ?


    3) \sum_{n=0}^{ \infty} \frac{1}{n!}

    My solution:
    b_n = \frac{n}{n!} In the Ratio test Converges

    \frac{1}{n!} < \frac{n}{n!}
    Then:

    \sum_{n=0}^{ \infty} \frac{1}{n!} is convergence

    Is correct ?



    4) \sum_{n=1}^{\infty} \frac{(n+1)}{n^3}
    How can I resolve this?
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