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How do you prove: $$ \lim_{n\to\infty}\frac{(2n-1)!!}{(2n)!!}=0 $$ where $(2n)!!=2n(2n-2)(2n-4)\cdots6\cdot 4\cdot2.$

I would do the trick $$2n=\frac{(2n-1)+(2n+1)}{2}\ge\sqrt{2n-1}\cdot\sqrt{2n+1}.$$ Therefore we have:

$$ 0\le\frac{(2n-1)!!}{(2n)!!}\le\frac{(2n-1)\cdot(2n-3)\cdots}{\sqrt{2n+1}\cdot\sqrt{2n-1}\cdot\sqrt{2n-1}\cdot\sqrt{2n-3}\cdots}=\frac{1}{\sqrt{2n+1}}\xrightarrow{n\to\infty} 0 .$$

But I think this is somewhat tricky. What are other ways to prove this limit? Thanks.

Bach
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    see (https://math.stackexchange.com/questions/1334359/determine-convergence-of-the-series-sum-n-1-infty-frac2n-12n?noredirect=1&lq=1). actually the corresponding series diverges but Mark Viola answer really helps you! – Chinnapparaj R Jul 06 '19 at 00:59

2 Answers2

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A brute force solution would be to rewrite everything with proper factorials and then use Stirling's approximation.

A neater one would consist in taking the log:

$$\ln\left(\frac{(2n-1)!!}{(2n)!!}\right)=\sum \ln\left(\frac{2k-1}{2k}\right)=\sum \ln\left(1-1/2k\right)$$

Then using the fact that $\ln\left(1-1/2k\right)\sim -1/2k$ and that the harmonic series diverges, you may conclude that the log goes to $-\infty$, and that your limit is $0$.

FXV
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Another way to do it. $$a_n=\frac{(2n-1)!!}{(2n)!!}\implies \log(a_n)=\log((2n-1)!!)-\log((2n)!!)$$

Using the less used Stirling like approximation $$\log(p!!)=\frac{1}{4} \log \left(\frac{\pi }{2}\right) \cos (\pi p)+\frac{1}{2} ( \log (p)-1) p+\frac{1}{4} \left(\log (2 \pi )+2 \log \left({p}\right)\right)+\frac{1}{6 p}+O\left(\frac{1}{p^3}\right)$$ Using it twice and continuing with Taylor series gives $$\log(a_n)=\frac{1}{2} \log \left(\frac{1}{\pi n}\right)-\frac{1}{8 n}+O\left(\frac{1}{n^3}\right)$$ making $$a_n \sim \frac{e^{-\frac{1}{8 n}}}{\sqrt{\pi n} }$$

For illustration purposes $S_5=\frac{63}{256}\approx 0.246094$ while the approximation gives $\frac{1}{\sqrt[40]{e} \sqrt{5 \pi }}\approx 0.246084$.

Claude Leibovici
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