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Math Help - Help testing these series with any root test or ratio test.

  1. #1
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    Help testing these series with any root test or ratio test.

    Problem 1 Evaluate


    Σ ( 100^n (2n)! / (3n)! )
    n=1

    Converges or diverges?

    And,

    Problem 2 Evaluate


    Σ [(n^3 + n)/(2n^3 + 5n^2 − 3n + 1)]^(n+1)
    n=1

    Converges or diverges?
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  2. #2
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    nm
    Last edited by Zocken; June 5th 2009 at 08:39 AM.
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Zocken View Post
    1. \frac{100^2}{(n+1)!}*\frac{n!}{100^2}

    =\frac{n!}{(n+1)!}=\frac{1}{n+1}\rightarrow 0

    so converges. might want to check me on the simplification of the permutations, not exactly sure on that.
    I get the following using the ratio test:

     \lim_{n \to \infty} \Big| \frac {100^{n+1}(2n+2)!}{(3n+3)!} \frac {(3n)!}{(2n)!100^{n}} \Big|  = \lim_{n \to \infty} \frac {100(2n+2)(2n+1)}{(3n+3)(3n+2)(3n+1)} = 0 since the the numerator is a quadratic and the denominator is a cubic
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  4. #4
    MHF Contributor Amer's Avatar
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    the second one
    use root test

    \left(\left(\frac{n^3+n}{2n^3+5n^2-3n+1}\right)^{n+1}\right)^{\frac{1}{n+1}}

    lim_{n\rightarrow \infty}\frac{n^3+n}{2n^3+5n^2-3n+1}=\frac{1}{2}<1...converge
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  5. #5
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    nm
    Last edited by Zocken; June 5th 2009 at 08:39 AM.
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  6. #6
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    Hey thanks all for your help.

    Random Variable:
    There is a part that i dont understand which is when you went from





    What happened with the (3n)! and the (2n)!?
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  7. #7
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by tapiaghector View Post
    Hey thanks all for your help.

    Random Variable:
    There is a part that i dont understand which is when you went from





    What happened with the (3n)! and the (2n)!?
    lim_{n\rightarrow\infty}\left(\frac{100^{n+1}(2n+2  )!}{(3n+3)!}\frac{(3n)!}{(2n)!100^{n}}\right)


    lim_{n\rightarrow\infty}\left(\frac{100^{n+1}(2n+2  )(2n+1)(2n)!}{(3n+3)(3n+2)(3n+1)(3n)!}\frac{(3n)!}  {(2n)!100^{n}}\right)

    as you can see (2n)! divide (2n)! =1 the same thing to (3n)!
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  8. #8
    Super Member Random Variable's Avatar
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    Quote Originally Posted by tapiaghector View Post
    Hey thanks all for your help.

    Random Variable:
    There is a part that i dont understand which is when you went from





    What happened with the (3n)! and the (2n)!?
     \frac {100^{n+1} (2n+2)(2n+1)(2n)!}{(3n+3)(3n+2)(3n+1)(3n)!} \frac {(3n)!}{(2n!) 100^{n}}
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