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Math Help - how do you simplify this series and test for convergence?

  1. #1
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    Angry how do you simplify this series and test for convergence? IMAGE-SOLUTION included..

    I have the series Σ from k=1 to infinity, [(k!)^2]/(2k)!

    I don't even know how to perform a ratio test cause I don't have any clue how i could simplify this... :/
    Last edited by nappysnake; December 6th 2011 at 08:20 AM.
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  2. #2
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    Re: how do you simplify this series and test for convergence?

    \lim_{k \to \infty}\frac{(k!)^2/(2k)!}{(k+1!)^2/(2k+2)!}=\lim_{k \to \infty}\frac{(2k+1)(2k+2)}{(k+1)(k+1)}=4

    convergent
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  3. #3
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    Re: how do you simplify this series and test for convergence?

    Quote Originally Posted by nappysnake View Post
    I have the series Σ from k=1 to infinity, [(k!)^2]/(2k)!
    I don't even know how to perform a ratio test cause I don't have any clue how i could simplify this... :/
    Consider the following:
    \begin{align*}(2n)! &=(2n)(2n-1)(2n-2)\cdots(2)(1)\\ &=2^n(n)(n-1)\cdots(1)(2n-1)(2n-3)\cdots(1)\\ &=2^n(n!)(2n-1)(2n-3)\cdots(1)   \end{align*}

    Now \frac{(n!)^2}{(2n)!}=\frac{n!}{2^n(2n-1)(2n-3)\cdots(3)(1)}
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  4. #4
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    Re: how do you simplify this series and test for convergence?

    wnvl and Plato, thank you, you helped me come to some realizations!
    Last edited by nappysnake; December 6th 2011 at 11:29 AM.
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