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Find $$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}.$$

I have tried the following: $$(2n-1)!!=\frac{(2n)!}{2^{2n}n!}$$ $$(2n)!!=2^nn!$$ $$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}=\lim_{n\rightarrow \infty}\frac{(2n)!}{2^{2n}(n!)^2}$$

Now using Stirling approximation, $$n!=\sqrt {2\pi n}\left(\frac{n}{e}\right)^{n}$$ $$(2n)!=\sqrt {2\pi 2n}\left(\frac{2n}{e}\right)^{2n}$$

$$\lim_{n\rightarrow \infty}\frac{(2n)!}{2^{2n}(n!)^2}=\lim_{n\rightarrow \infty}\frac{\sqrt{2\pi}\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{2^{2n}(\sqrt{2\pi})^2(\sqrt{n})^2\left(\frac{n}{e}\right)^{2n}}$$ $$=\frac{1}{\sqrt{2\pi}}\lim_{n\rightarrow \infty}\frac{\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{2^{2n}(\sqrt{n})^2\left(\frac{n}{e}\right)^{2n}}$$

How to proceed with solving this limit?

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  • $\begingroup$ Open the parentheses to get an expression of the form $\lim_{n\to\infty}c\cdot n^\alpha$. Then it should be easy. $\endgroup$
    – Guy
    Nov 4, 2017 at 13:55
  • $\begingroup$ $$(2n/e)^{2n} = 2^{2n} \cdot (n/e)^{2n}$$ $\endgroup$ Nov 4, 2017 at 13:55
  • $\begingroup$ You've almost finished $\endgroup$
    – Elnur
    Nov 4, 2017 at 14:08
  • $\begingroup$ Note that after $$\lim_{n\rightarrow \infty}\frac{(2n)!}{2^{2n}(n!)^2}$$ you can instead move to $$\lim_{n \to \infty} \binom{2n}{n} \times \frac{1}{2^{2n}}$$ which is the fraction of the $n$-dimensional unit cube that is on the middle "layer"; so this actually has a combinatorial application: it shows that "exponentially little of the unit cube is actually near the centre", which is nonrigorously implied already by the fact that the unit cube is of volume $2^n$ but the unit sphere only of volume $1$. $\endgroup$ Nov 4, 2017 at 14:17
  • $\begingroup$ Another post about the same limit: How to prove that $\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0$. And also other questions linked there. $\endgroup$ Apr 13, 2019 at 13:32

3 Answers 3

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We have $$ \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^n}\binom{2n}{n} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)=\frac{1}{2}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right) \tag{A}$$ and by squaring both sides $$ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)=\frac{1}{4n}\prod_{k=1}^{n-1}\left(1+\frac{1}{4k(k+1)}\right)\tag{B} $$ from which: $$ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\leq \frac{1}{4n}\prod_{k=1}^{n-1}\exp\left(\frac{1}{4k}-\frac{1}{4(k+1)}\right)\leq \frac{e^{1/4}}{4n}\tag{C} $$ implying that the wanted limit is zero.

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$$\frac{1}{\sqrt{2\pi}}\lim_{n\rightarrow \infty}\frac{\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{2^{2n}(\sqrt{n})^2\left(\frac{n}{e}\right)^{2n}}=\frac{1}{\sqrt{2\pi}}\lim_{n\rightarrow \infty}\frac{\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{(\sqrt{n})^2\left(\frac{2n}{e}\right)^{2n}} =\frac{1}{\sqrt{2\pi}}\lim_{n\rightarrow \infty}\frac{\sqrt{2n}}{(\sqrt{n})^2}$$ $$=\frac{1}{\sqrt{\pi}}\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}$$

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We have $$ \frac{(2n-1)!!}{(2n)!!} = \frac{(2n-1)!!}{((2n)(2n-1))!!}\le \frac{(2n-1)!!}{(2n)((2n-1)!!)}=\frac{1}{2n}$$ Now we have upper bound on our equation. Note that as $n \to \infty$, our upper bound tends to 0. $$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!} \le0$$

Hence our answer is 0

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  • $\begingroup$ It's a bit unclear to me what you meant in this step: $\frac{(2n-1)!!}{(2n)!!} = \frac{(2n-1)!!}{((2n)(2n-1))!!}$. If you wanted to write $\frac{(2n-1)!!}{(2n)!!} = \frac{(2n-1)!!}{(2n)((2n-2))!!}$, then you get into problems with your next inequality. $\endgroup$ Apr 13, 2019 at 9:51

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