What are common lower bounds for ${{2n} \choose {n}}$?
Edit: I made a mistake in my original question.
It doesn't change my question but there is no reason for me to include the mistake.
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You could try stirling's approximation on $\frac{(2n)!}{n!^2}$ – Mar 20 '18 at 19:34
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Stirling's gives roughly $4^n$ I do not recall the details, but easy enough to do again: – Will Jagy Mar 20 '18 at 19:34
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1$2n/n=2$, not $n$. – John Brevik Mar 20 '18 at 19:34
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1Here is a tight lower bound for Catalan numbers, which are $\frac{1}{n+1} {2n \choose n}$. – mechanodroid Mar 20 '18 at 19:35
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Using the bounds from this answer, we have $$ \frac{4^n}{\sqrt{\pi\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\left(n+\frac14\right)}} $$
robjohn
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good; from Stirling's, the squared denominator keeps an extra $\sqrt n$ compared with the denominator. I wasn't sure yet – Will Jagy Mar 20 '18 at 19:44
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1@gimusi: you can easily notice that $\frac{1}{\sqrt{\pi}}>\frac{2\sqrt{\pi}}{e^2}$, hence robjohn's bound is sharper than yours. Also because the constant in robjohn's answer is optimal. – Jack D'Aurizio Mar 20 '18 at 20:07
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@JackD'Aurizio oh yes of course! Indeed it is a very narrow bound. I will take a look to the derivation. Thanks – user Mar 20 '18 at 20:09
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@Jack D'Aurizio In wich sense is the constant optimal? Is 1/3 the smallest posible real number, such that the inequality holds for all $n$? – user Mar 20 '18 at 21:03
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@user: I meant that the constant $\frac{1}{\sqrt{\pi}}$ is optimal, since $$ \lim_{n\to +\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}=\frac{1}{\sqrt{\pi}}.$$ Notice that $\sqrt{\pi}=\Gamma\left(\frac{1}{2}\right)$, and indeed the identity above follows from Legendre's duplication formula for the $\Gamma$ function. – Jack D'Aurizio Mar 20 '18 at 21:12
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1@user: note that $\lim\limits_{n\to\infty}\left[,\raise{4pt}{\frac{16^n}{\pi\binom{2n}{n}^2}}-n,\right]=\frac14$, so the $\frac14$ in the upper bound is optimal. The $\frac13$ in the lower bound can be replaced by $\frac1\pi$ and the inequality will still be valid for $n\ge0$. – robjohn Mar 20 '18 at 21:29
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Using Stirling's bound
$$\sqrt{2\pi}\ n^{n+\frac12}e^{-n} \le n! \le e\ n^{n+\frac12}e^{-n}$$
we obtain
$$\binom{2n}{n}=\frac{(2n)!}{n!^2}\ge\frac{\sqrt{2\pi}\ (2n)^{2n+\frac12}e^{-2n}}{e^2\ n^{2n+1}e^{-2n}}=\frac{\sqrt{2\pi}\ 2^{2n}2^{\frac12}n^{2n+\frac12}}{e^2\ n^{2n+1}}=\frac{2\sqrt{\pi}}{e^2}\frac{4^n}{\sqrt n}$$
user
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Erdös had a really nice lower bound, which he used in his proof of Bertrand's Postulate: $4^n/2n \leqslant {{2n}\choose{n}}$. This follows because \begin{align*} (1+1)^{2n}=\sum_{k=0}^{2n} {{2n}\choose{k}} < 1+2n{{2n}\choose{n}} \end{align*} and then $4^n \leqslant 2n{{2n}\choose{n}}$. In fact the bound can be strengthened without too much difficulty to give $4^n/n \leqslant {{2n}\choose{n}}$.
Edit: The stronger bound only holds for $n \geqslant 4$.
primes.against.humanity
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