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Math Help - [SOLVED] Limits question tuffy

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

    any ideas how 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 \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 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 \frac{(n+1)^2(2n)!}{(2n+2)!} ?? thnx
    So  \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)}  = \lim_{n \to \infty} \frac{n^2+2n+1}{(2n+2)(2n+1)}   . Note that the highest degree term in the numerator is  n^2 and the highest degree term in the denominator is  4n^2 . So as  n \to \infty ,  \frac{n^2}{4n^2} \to \frac{1}{4} .
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  3. #3
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    Try to use these:
    \begin{gathered}<br />
  \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 \\<br />
  \left( {2n + 2} \right)! = \left( {2n + 2} \right)\left( {2n + 1} \right)\left( {2n} \right)! \hfill \\ <br />
\end{gathered}
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