I meet an expression as following: $$L_{N}=\frac{C_{2N}^{N}}{2^{2N}}=\frac{(2N)!}{N!N!2^{2N}}$$
My conjecture is that: $$\lim_{N \to \infty}L_{N}=0$$
It seems to be an easy question, but how to show it simply?
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$\begingroup$ Use the aymptotic of the Catalan numbers to get that the fraction behaves like $(N+1)/N^{3/2}$. $\endgroup$– DirkCommented Dec 14, 2016 at 15:42
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$\begingroup$ If $k\approx N$, then $C_{2N}^k\approx C_{2N}^N$. $\endgroup$– Ilya BogdanovCommented Dec 14, 2016 at 15:42
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$\begingroup$ Note : $L_n$ is the ratio of the product of the first $n$ odd numbers to the product of the first $n$ even numbers. Possibly, that helps. $\endgroup$– Aditya Guha RoyCommented Sep 12, 2017 at 17:39
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2 Answers
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Erdos gives ${{2n}\choose{n}}4^{-n}<1/\sqrt{2n+1}$ in his book with Suranyi. For an inductive proof one needs to show $$ \frac{2n(2n-1)}{4n^2}\cdot\frac{1}{\sqrt{2n-1}}<\frac{1}{\sqrt{2n+1}} $$ which simplifies to $\sqrt{2n-1}\sqrt{2n+1}<2n$.
answered Dec 14, 2016 at 16:11
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$\begingroup$ Komjath : could you please mention the name of the book. $\endgroup$ Commented Sep 12, 2017 at 17:40
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1$\begingroup$ Yes: Topics in the Theory of Numbers, springer.com/gp/book/9780387953205 See also at google.books books.google.hu/books/about/… $\endgroup$ Commented Sep 15, 2017 at 4:18
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You're right. It is easy with Stirling's approximation $$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$ so that $$L_N=\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\sqrt{\pi N}}\rightarrow 0$$ as $N\rightarrow\infty$.
answered Dec 14, 2016 at 15:46
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$\begingroup$ Thanks very much! Stirling's approximation is so powerful! $\endgroup$– GalorCommented Dec 15, 2016 at 1:48