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Math Help - infinite series

  1. #1
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    infinite series

    so if I use the comparison test to determine whether the infinite series, \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{2n^3 + 7n}}, and using \frac{1}{n^{\frac{3}{2}}} as the comparing series....should the initial infinite series be convergent or divergent

    my belief is that it should be divergent because \frac{1}{n^{\frac{3}{2}}} is divergent and it is greater than \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{2n^3 + 7n}}

    Thanks in advance for any help provided
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  2. #2
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    Response

    This is what I found

    The p-series is given by
    Σ[1/(n^p)] = 1/(1^p) + 1/(2^p) + 1/(3^p) + ...
    where p > 0 by definition.
    If p > 1, then the series converges.
    If 0 < p <= 1 then the series diverges.

    Since 3/2 > 1, the series should converge. Hope this helps.
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  3. #3
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    Quote Originally Posted by ajj86 View Post
    This is what I found

    The p-series is given by
    1/(n^p) = 1/(1^p) + 1/(2^p) + 1/(3^p) + ...
    where p > 0 by definition.
    If p > 1, then the series converges.
    If 0 < p <= 1 then the series diverges.

    Since 3/2 > 1, the series should converge. Hope this helps.

    AH~ the p-series. good thinking 99.
    So in effect the initial series converges as well....correct?
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  4. #4
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    Response

    You got it. Basically, your statement is correct if you substitute convergent for divergent in the sentence you wrote above:
    my belief is that it should be convergent because <br />
\frac{1}{n^{\frac{3}{2}}}<br />
    is convergent and it is greater than

    <br />
\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{2n^3 + 7n}}<br />
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