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  1. #1
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    Series

    Is this series absolutely convergent

    2/5 + (2*6)/(5*8) + (2*6*10)/(5*8*11) + (2*6*10*14)/(5*8*11*14) + ...
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by amiv4 View Post
    Is this series absolutely convergent

    2/5 + (2*6)/(5*8) + (2*6*10)/(5*8*11) + (2*6*10*14)/(5*8*11*14) + ...
    \sum_{n=0}^{\infty}\frac{2\cdot{6}\cdot{10}\cdots{  4n+2}}{5\cdot{8}\cdot{11}\cdot{3n+5}}=\sum_{n=0}^{  \infty}\frac{\prod_{k=0}^{n}(4k+2)}{\prod_{k=0}^{n  }(3k+5)}

    Using the Ratio Test givse

    \begin{aligned}\lim_{n\to\infty}\left|\frac{\prod_  {k=0}^{n+1}(4k+2)}{\prod_{k=0}^{n+1}(3k+5)}\cdot\f  rac{\prod_{k=0}^{n}(3k+5)}{\prod_{k=0}^{n}(4k+2)}\  right|&=\lim_{n\to\infty}\left|\frac{4(n+1)+2}{3(n  +1)+5}\right|\\<br />
&=\frac{4}{3}>1<br />
\end{aligned}

    Therefore your series diverges.
    Last edited by Mathstud28; November 11th 2008 at 12:16 PM. Reason: typo
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  3. #3
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    shouldnt it be 4k +2 in the first part instead of 2k+4. and i think that messes everything else up
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by amiv4 View Post
    shouldnt it be 4k +2 in the first part instead of 2k+4. and i think that messes everything else up
    Yes, that was a typo. The solution still holds.
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  5. #5
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    k I'm really not seeing how when u multiply those two together you get that. Cuz i did the ratio test myself and i dont see how those multiply together to get that. Can you explain
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  6. #6
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    Let's write it explicitly.

    Let: a_{n} = \frac{2\cdot{6}\cdot{10}\cdots{4n+2}}{5\cdot{8}\cd  ot{11}\cdots{3n+5}}

    Using the ratio test,
    \begin{aligned} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}}\right| & = \lim_{n \to \infty} \left|  \frac{2\cdot{6}\cdot{10}\cdots{4(n+1)+2}}{5\cdot{8  }\cdot{11}\cdots{3(n+1)+5}} \ \cdot \  \frac{5\cdot{8}\cdot{11}\cdots{3n+5}}{2\cdot{6}\cd  ot{10}\cdots{4n+2}} \right| \end{aligned} (simply what \frac{a_{n+1}}{a_{n}} is)

    {\color{white}\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}}\right|}  = \lim_{n \to \infty} \left| \frac{2\cdot{6}\cdot{10}\cdots{4n+6}}{5\cdot{8}\cd  ot{11}\cdots{3n+8}} \ \cdot \ \frac{5\cdot{8}\cdot{11}\cdots{3n+5}}{2\cdot{6}\cd  ot{10}\cdots{4n+2}} \right|<br />
(simplified the brackets)

     {\color{white}\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}}\right|} = \lim_{n \to \infty} \bigg| \frac{{\color{red} 2\cdot{6}\cdot{10}\cdots (4n+2)}(4n+6)}{{\color{blue}5\cdot{8}\cdot{11}\cdo  ts (3n+5)}(3n+8)} {\color{white}.} \ \cdot \  \frac{{\color{blue}5\cdot{8}\cdot{11}\cdots{3n+5}}  }{{\color{red}2\cdot{6}\cdot{10}\cdots{4n+2}}} \bigg| (expanded the inside of the dots)

    Notice that the red's and blue's cancel out. Take the limit of what remains and the rest follows (see Mathstud's post).
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