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This question already has an answer here:

How to prove $$\frac{1}{4^n}\binom{2n}{n}\leq \frac{1}{\sqrt{\pi n}} ?$$

What i tried:

$$\frac{1}{4^n}\frac{(2n)!}{n!\times n!} = \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \frac{2n-1}{2n}.$$

Arithmetic inequality

$$\frac{2r-1+2r+1}{2}\geq \sqrt{(2r-1)(2r+1)}$$

$$2r\geq \sqrt{(2r-1)(2r+1)}$$

$$\frac{1}{2r}\leq \sqrt{\frac{1}{(2r-1)(2r+1)}}$$

$$\prod^{n}_{r=1}\frac{2r-1}{2r}\leq \prod^{n}_{r=1}\sqrt{\frac{2r-1}{2r+1}}=\frac{1}{\sqrt{2n+1}}\;\;, n\geq 1,$$

but how do I prove my original inequality for natural numbers? Help me, please!

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marked as duplicate by Arnaud D., awkward, Community Feb 9 at 8:21

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  • $\begingroup$ That upper bound seems to come from Stirling's formula applied to $n!$ and $(2n)!$. $\endgroup$ – PierreCarre Feb 5 at 10:04
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May be not the way you want, but let $$a_n=\frac{1}{4^n}\binom{2n}{n}=\frac{1}{4^n}\frac{(2n)!}{n!\times n!}$$ Take logarithms $$\log(a_n)=\log((2n)!)-2\log(n!)-n\log(4)$$ Use Stirling approximation for the factorials and continue with Taylor to get $$\log(a_n)=-\frac{1}{2} \log \left({\pi n}\right)-\frac{1}{8 n}+O\left(\frac{1}{n^3}\right)$$ making $$a_n\simeq \frac 1{\sqrt{\pi n}}e^{-\frac 1 {8n}} < \frac 1{\sqrt{\pi n}}$$

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I would rather used the following result : $$\left|\left(\frac{\mathrm e}n\right)^n\,n!-\sqrt{2\pi n}\right| \leq 2$$

Using it as an upper bound for $2n!$ and a lower bound for $n!$:

$$ \binom{2n}{n} \leq \frac{(2\sqrt{\pi n}+2)\left(\frac{2n}{e}\right)^{2n}}{(\sqrt{2\pi n}-2)\left(\frac{n}{e}\right)^{n}}$$

Leading to the desired result

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