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Show that $\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} \left(1+\mathcal{O}\bigl(\frac{1}{n}\bigr)\right)$.

Hint: It might be useful that for $f,g: \mathbb{N} \rightarrow \mathbb{R}_{\ge 0}$ with $f(n) = \mathcal{O}(g(n))$ and $g(n) = \mathcal{o}(1)$ holds $\frac{1}{1+f(n)} = 1 +\mathcal{O}(g(n))$.

I tried to evaluate the limit superior of

$$ \bigg\lvert \frac{\binom{2n}{n} - \frac{4^n}{\sqrt{\pi n}} }{\frac{1}{n}} \bigg\rvert = \bigg\lvert \frac{2n \cdots (n+1)}{(n-1)!} - \frac{4^n\sqrt{n}}{\sqrt{\pi}} \bigg\rvert$$

, but I do not get farther than that. Could you please give me a hint?

Thomas Andrews
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3nondatur
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    I don't think I know a way to do this one without Stirling. – Ian Dec 01 '22 at 16:35
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    Does this answer your question? Elementary central binomial coefficient estimates – Anne Bauval Dec 01 '22 at 16:43
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    I think as written that error can't be right? I think you want $(1 + \mathcal{O}(1/n))$? – Jakob Streipel Dec 01 '22 at 16:52
  • Yeah, seems unlikely to be true, but maybe, as prets suggests: $$\binom{2n}n=\frac{4^n}{\sqrt{\pi n}}\left(1+\mathcal O\left(\frac 1n\right)\right)$$ – Thomas Andrews Dec 01 '22 at 16:56
  • Basically, the reason this seems unlikely to be true is that we use Stirling's formula to prove this, using $$n!=\sqrt{2\pi n}\left(\frac ne\right)^n\left(1+\mathcal O\left(\frac 1n\right)\right)$$ https://en.wikipedia.org/wiki/Stirling%27s_approximation – Thomas Andrews Dec 01 '22 at 17:02
  • Sorry, I made a typo and corrected it – 3nondatur Dec 01 '22 at 17:04
  • But now your limit superior method would never get the right result. Now you need the limit superior of $$n\left|\left(\binom{2n}{n}\frac{\sqrt{\pi n}}{4^n}-1\right)\right|$$ – Thomas Andrews Dec 01 '22 at 17:06
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    If you don't already know Stirling's approximation, the weird part is the $\sqrt{\pi n},$ which might seem to jump out of nowhere. – Thomas Andrews Dec 01 '22 at 17:09
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    But the hint would seem to imply using Stirling. Then you get: $$\binom{2n}n=\frac{\sqrt{4\pi n}\left(\frac {2n}e\right)^{2n}\left(1+O\left(\frac1{2n}\right)\right)}{2\pi n\left(\frac ne\right)^{2n}\left(1+O\left(\frac 1n\right)\right)^2}=\dfrac{4^n}{\sqrt{\pi n}}\dfrac{\left(1+O\left(\frac1{2n}\right)\right)}{\left(1+O\left(\frac 1n\right)\right)^2}$$ The hint seems to be to help you resolve that right part - how to show: $$\dfrac{\left(1+O\left(\frac1{2n}\right)\right)}{\left(1+O\left(\frac 1n\right)\right)^2}=1+O\left(\frac 1n\right)$$ – Thomas Andrews Dec 01 '22 at 17:18
  • @ThomasAndrews: I understand that $\mathcal{O}(\frac{1}{2n}) = \mathcal{O}(\frac{1}{n})$, but I am not sure if we can use cancellation law for fractions containing big Ohs. – 3nondatur Dec 01 '22 at 18:43
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    @3nondatur You can't just cancel, since you could have $(1+1/n)/(1+2/n)$ for example. But there is still something you can do instead to show the relation Thomas Andrews wrote... – Ian Dec 01 '22 at 19:06
  • @Ian: Sorry, but I do not get it. – 3nondatur Dec 01 '22 at 20:18
  • Hint: Firs show $(1+O(1/n))^2=1+O(1/n).$ Then show: $$\frac{1+O(1/n)}{1+o(1/n)}=1+O(1/n)$$ – Thomas Andrews Dec 01 '22 at 20:58
  • @ThomasAndrews: Thanks a lot, I get it now. – 3nondatur Dec 01 '22 at 21:12

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