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Thread: Series: comparison test

  1. #1
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    Series: comparison test

    Use the comparison test to investigate the convergence or divergence

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

    My solution:

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

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


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

    Is correct ?


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

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

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

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

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

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

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

    It is correct ?


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

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

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

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

    Is correct ?



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