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Compute $$\lim_{n\rightarrow\infty}\left(\frac{\left(2n\right)!}{\left(n!\right)^{2}}\right)^{\frac{1}{n}}$$ If you have some nice proofs and you're willing to share them, then I thank you and you definitely have my upvote!

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Use \$\$ for tex commands –  Kaster Feb 12 '13 at 8:12
    
done. thank you! –  tomer Feb 12 '13 at 8:14

3 Answers 3

up vote 4 down vote accepted

Note that $$ \frac{(2n)!}{(n!)^2}=\binom{2n}{n} $$ Then, using the simple identity $$ \binom{2n+2}{n+1}=\binom{2n}{n}\frac{4n+2}{n+1} $$ by induction, we get that $$ \binom{2n}{n}=4^n\prod_{k=0}^{n-1}\frac{k+{\small1/2}}{k+1} $$ Since $$ \frac1{2n}=\underbrace{\frac12}_{k=0}\underbrace{\left(\frac12\cdot\frac23\cdot\frac34\cdots\frac{n-1}n\right)}_{1\le k\le n-1}\le\prod_{k=0}^{n-1}\frac{k+{\small1/2}}{k+1}\le1 $$ Therefore, $$ \frac1{2n}4^n\le\binom{2n}{n}\le4^n $$ Thus, by the Squeeze Theorem $$ \lim_{n\to\infty}\binom{2n}{n}^{1/n}=4 $$

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You can use the Stirling approximation, $$n!\sim n^n e^{-n}\sqrt{2\pi n}$$ from which we have $$\left(\frac{(2n)!}{(n!)^2}\right)^{1/n}\sim \left(\frac{(2n)^{2n}e^{-2n}\sqrt{4\pi n}}{n^{2n} e^{-2n}2\pi n}\right)^{1/n}=4\cdot\frac{1}{(\pi n)^{1/(2n)}}.$$ The latter term can be written $$\frac{1}{(\pi n)^{1/(2n)}}=\exp\left(-\frac{1}{2n}\ln(\pi n)\right)\to 1,$$ so the limit is $4$.

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You could use the fact that for a sequence $(a_n)$ of positive numbers, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty} {\root n\of{a_n}}$ and the two limits are equal. For a proof of this, see these notes of Pete L. Clark.

Apply this result to $a_n={(2n)!\over (n!)^2}$. One easily evaluates $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}=4$. Then $\lim\limits_{n\rightarrow\infty} {a_n^{1/n}}=4$.

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This is a nice result. –  Mhenni Benghorbal Feb 12 '13 at 13:58

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