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Math Help - Ratio Test on a Series

  1. #1
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    Ratio Test on a Series

    The series is \Sigma_{n=0}^{\infty}\frac{(n!)}{(3n)!}

    \lim_{n->\infty}\mid\frac{a_{n+1}}{a_n}\mid<1 \rightarrow \Sigma a_n converges. If it's less than 1, then the series diverges; if it's equal to 1, then the test is inconclusive.

    \lim_{n->\infty}\left(\frac{[(n+1)!]^2}{[3(n+1)]!}\right)\left(\frac{(3n)!}{(n!)^2}\right)

    =\lim_{n->\infty}\frac{(n+1)(3n)!}{[3(n+1)]!}

    I can't seem to simplify this any further.
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  2. #2
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    Quote Originally Posted by adkinsjr View Post
    The series is \Sigma_{n=0}^{\infty}\frac{(n!)}{(3n)!}

    \lim_{n->\infty}\mid\frac{a_{n+1}}{a_n}\mid<1 \rightarrow \Sigma a_n converges. If it's less than 1, then the series diverges; if it's equal to 1, then the test is inconclusive.

    \lim_{n->\infty}\left(\frac{[(n+1)!]^2}{[3(n+1)]!}\right)\left(\frac{(3n)!}{(n!)^2}\right)

    =\lim_{n->\infty}\frac{(n+1)(3n)!}{[3(n+1)]!}

    I can't seem to simplify this any further.


    Where did you take the squares from in your third line? Anyway, \frac{(3n)!}{(3n+3)!}=\frac{1}{(3n+1)(3n+2)(3n+3)} and from here you get at once that the limit is zero and thus the series converges.

    Tonio
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