[SOLVED] Limits question tuffy

**i_zz_y_ill** · May 29th 2009, 03:24 PM

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

**Sampras** · May 29th 2009, 03:37 PM

Originally Posted by i_zz_y_ill

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 $n \to \infty$ , $\frac{n^2}{4n^2} \to \frac{1}{4}$ .

**Plato** · May 29th 2009, 03:39 PM

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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