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Math Help - series limit RP

  1. #1
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    series limit RP

     lim_{n \to \infty} \frac{1*3*5*....*(2n-1)}{2*4*6*...*(2n)} = ?
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  2. #2
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    Mistaken post. I can't remove it.
    Last edited by Sudharaka; February 2nd 2010 at 04:44 PM.
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  4. #4
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    (2)(4)(6)...(2n-2)(2n)= (2*1)(2*2)(2*3)...(2*(n-1))(2*n)= 2^n n!.

    1(3)(5)...(2n-1)= \frac{1(2)(3)(4)...(2n-1)(2n)}{2(4)(6)...(2n)}= \frac{(2n)!}{2^n n!} so

    \frac{1*3*5*\cdot\cdot\cdot(2n-1)}{2*4*6*\cdot\cdot\cdot (2n)}= \frac{(2n)!}{2^{2n}n!}

    That 2^{2n} in the denominator "dominates" is enough to show that the limit is 0.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    (2)(4)(6)...(2n-2)(2n)= (2*1)(2*2)(2*3)...(2*(n-1))(2*n)= 2^n n!.

    1(3)(5)...(2n-1)= \frac{1(2)(3)(4)...(2n-1)(2n)}{2(4)(6)...(2n)}= \frac{(2n)!}{2^n n!} so

    \frac{1*3*5*\cdot\cdot\cdot(2n-1)}{2*4*6*\cdot\cdot\cdot (2n)}= \frac{(2n)!}{2^{2n}n!}

    That 2^{2n} in the denominator "dominates" is enough to show that the limit is 0.
    Dear HallsofIvy,

    I think you have made a slight error. The last line must be,
    \frac{1*3*5*\cdot\cdot\cdot(2n-1)}{2*4*6*\cdot\cdot\cdot (2n)}= \frac{(2n)!}{2^{2n}(n!)^2}
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