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Thread: Limit question

  1. #1
    Senior Member Twig's Avatar
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    Limit question

    Hi

    Calculate the limit:

    $\displaystyle \displaystyle\lim_{n\to\infty} \binom{2n}{n} $


    Where do I go wrong?

    $\displaystyle \binom{2n}{n} = \frac{2n!}{n!(2n-n)!} = \frac{2^{n}\cdot n!}{n!\cdot n!} = \frac{2^{n}}{n!}$

    Thx
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    $\displaystyle (2n)!\ =\ 2^n\cdot n!\cdot1\cdot3\cdot5\cdot\cdots\cdot(2n-1)$
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  3. #3
    Senior Member Twig's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    $\displaystyle (2n)!\ =\ 2^n\cdot n!\cdot1\cdot3\cdot5\cdot\cdots\cdot(2n-1)$

    I am not really with you on why this is true.
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    Okay. Let’s take 20! for example.

    $\displaystyle (2\times10)!$

    $\displaystyle =\quad1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\ cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cd ot16\cdot17\cdot18\cdot19\cdot20$

    $\displaystyle =\quad2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\cdo t16\cdot18\cdot20\cdot1\cdot3\cdot5\cdot7\cdot9\cd ot11\cdot13\cdot15\cdot17\cdot19$

    $\displaystyle =\quad2^{10}(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7 \cdot8\cdot9\cdot10)\cdot1\cdot3\cdot5\cdot7\cdot9 \cdot11\cdot13\cdot15\cdot17\cdot19$

    $\displaystyle =\quad2^{10}\cdot10!\cdot1\cdot3\cdot5\cdot7\cdot9 \cdot11\cdot13\cdot15\cdot17\cdot19$

    Yes?
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