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Math Help - series?convergent

  1. #1
    Member zangestu888's Avatar
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    series?convergent

    \sum_{n=0}^{\infty}3+sin(n)/n^3+2
    how would i do this? all i know is the sequence of partial sums is a finite vaule , 0? correct wouldnt that mean that my series will be convergent?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by zangestu888 View Post
    \sum_{n=0}^{\infty}3+sin(n)/n^3+2
    how would i do this? all i know is the sequence of partial sums is a finite vaule , 0? correct wouldnt that mean that my series will be convergent?
    If you mean

    \frac{3+\sin(n)}{n^3+2}

    Then by the comparison test

    \left| \frac{3+\sin(n)}{n^3+2} \right|< \frac{4}{n^3}

    The latter series is convergent by p-series or the integral test
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  3. #3
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    Hello, zangestu888!

    S \;=\;\sum_{n=0}^{\infty}\frac{3+\sin(n)}{n^3+2}
    I would use the Comparison Test.


    Since \sin(n) \:\leq\: 1, then: . 3 + \sin(n) \:\leq 4 .[1]

    Since n^3 + 2 \:>\: n^3, then: . \frac{1}{n^3+2} \:< \:\frac{1}{n^3} .[2]

    Multiply [1] and [2]: . \frac{3+\sin(n)}{n^3+2} \:< \frac{4}{n^3}

    Take sums: . S \;=\;\sum^{\infty}_{n=1}\frac{3+\sin(n)}{n^3+2} \;< \;\sum^{\infty}_{n=1}\frac{4}{n^3} \quad\Leftarrow\:\text{ a convergent }p\text{-series}

    Therefore, S converges.



    Too slow again . . . \emptyset beat me to it.
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