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Math Help - Determine whether the series is absolutely convergent, conditionally convergent ....

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    Determine whether the series is absolutely convergent, conditionally convergent ....

    Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

    Last edited by mr fantastic; April 25th 2010 at 08:41 PM. Reason: Re-titled post.
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    Quote Originally Posted by alin1916 View Post
    Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

    Ratio test:

    \lim_{k \to \infty}\left|\frac{a_{k + 1}}{a_k}\right|

     = \lim_{k \to \infty}\left|\frac{\frac{(-1)^{k + 2}\cdot 1 \cdot 4 \cdot 7 \cdot \dots \cdot (3k - 2) \cdot (3k + 1)}{(k + 1)!2^{k + 1}}}{\frac{(-1)^{k + 1}\cdot 1 \cdot 4 \cdot 7 \cdot \dots \cdot (3k - 2)}{k!2^k}}\right|

     = \lim_{k \to \infty}\frac{\frac{1 \cdot 4  \cdot 7 \cdot \dots \cdot (3k - 2) \cdot (3k + 1)}{(k + 1)!2^{k +  1}}}{\frac{1 \cdot 4 \cdot 7 \cdot \dots \cdot (3k -  2)}{k!2^k}}

     = \lim_{k \to \infty}\frac{3k + 1}{2(k + 1)}

     = \lim_{k \to \infty}\frac{3k + 1}{2k + 2}

     = \lim_{k \to \infty}\frac{3}{2} by L'Hospital's Rule

     = \frac{3}{2}

     > 1.


    So the series is divergent.
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