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Thread: [SOLVED] Limits question tuffy

  1. #1
    Member i_zz_y_ill's Avatar
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    [SOLVED] Limits question tuffy

    any ideas how $\displaystyle Lim_{n\rightarrow\infty}\frac{((n+1)!)^2(2n)!}{(2n +2)!(n!)^2}=Lim_{n\rightarrow\infty}\frac{(n+1)^2} {(2n+2)(2n+1)}=\frac{1}{4}$ I can only get as far as $\displaystyle \frac{(n+1)^2(2n)!}{(2n+2)!}$ ?? thnx
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  2. #2
    Senior Member Sampras's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    any ideas how $\displaystyle Lim_{n\rightarrow\infty}\frac{((n+1)!)^2(2n)!}{(2n +2)!(n!)^2}=Lim_{n\rightarrow\infty}\frac{(n+1)^2} {(2n+2)(2n+1)}=\frac{1}{4}$ I can only get as far as $\displaystyle \frac{(n+1)^2(2n)!}{(2n+2)!}$ ?? thnx
    So $\displaystyle \lim_{n \to \infty} \frac{[(n+1)!]^{2}(2n)!}{(2n+2)!(n!)^{2}} = \lim_{n \to \infty} \frac{[n! \cdot (n+1)]^{2} \cdot (2n)!}{(2n+1)! \cdot (2n+2) \cdot (n!)^{2}} = \lim_{n \to \infty} \frac{(n+1)^{2}}{(2n+2)(2n+1)} $ $\displaystyle = \lim_{n \to \infty} \frac{n^2+2n+1}{(2n+2)(2n+1)} $. Note that the highest degree term in the numerator is $\displaystyle n^2 $ and the highest degree term in the denominator is $\displaystyle 4n^2 $. So as $\displaystyle n \to \infty $, $\displaystyle \frac{n^2}{4n^2} \to \frac{1}{4} $.
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  3. #3
    MHF Contributor

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    Try to use these:
    $\displaystyle \begin{gathered}
    \left( {\left( {n + 1} \right)!} \right)^2 = \left( {\left( {n + 1} \right)n!} \right)^2 = \left( {n + 1} \right)^2 \left( {n!} \right)^2 \hfill \\
    \left( {2n + 2} \right)! = \left( {2n + 2} \right)\left( {2n + 1} \right)\left( {2n} \right)! \hfill \\
    \end{gathered} $
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