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Math Help - convergence of a series

  1. #1
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    convergence of a series

    determine convergence and show all work

    inf
    Sigma [3*5*7...(2n+1)] / [18^n(2n-1)n!]
    n=1


    I have been trying to do this problem for several days I have managed to figure out all the others but this one has me stumped.

    Thank Your for your time
    Kevin
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  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
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    802
    Hi

    Let's show the convergence using the ratio test. (it's often useful when the terms of the series are defined by fractions or products)

    \frac{a_{n+1}}{a_n}=a_{n+1}\cdot \frac{1}{a_n}=\frac{{\color{red}3\cdot 5\ldots(2n+1)}\cdot(2n+3)}{18^{n+1}(2n+1)(n+1)!}\c  dot \frac{18^n(2n-1)n!}{{\color{red}3\cdot5\ldots\cdot(2n+1)}}

    Simplifying what's in red and using (n+1)!=(n+1)\cdot n! and 18^{n+1}=18\cdot 18^n :

    \frac{a_{n+1}}{a_n}=\frac{2n+3}{18\cdot{\color{red  }18^n}(2n+1)\cdot (n+1)\cdot {\color{blue}n!}}\cdot {\color{red}18^n}(2n-1){\color{blue}n!}

    \frac{a_{n+1}}{a_n}=\frac{(2n+3)(2n-1)}{18(2n+1)(n+1)}

    The series converges if \lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right| <1.

    Hope that helps.
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