Lower bounds for ${{2n} \choose {n}}$

Question

What are common lower bounds for ${{2n} \choose {n}}$?

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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.

Answer 1

Using the bounds from this answer, we have $$ \frac{4^n}{\sqrt{\pi\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\left(n+\frac14\right)}} $$

Answer 2

Using Stirling's bound

$$\sqrt{2\pi}\ n^{n+\frac12}e^{-n} \le n! \le e\ n^{n+\frac12}e^{-n}$$

we obtain

$$\binom{2n}{n}=\frac{(2n)!}{n!^2}\ge\frac{\sqrt{2\pi}\ (2n)^{2n+\frac12}e^{-2n}}{e^2\ n^{2n+1}e^{-2n}}=\frac{\sqrt{2\pi}\ 2^{2n}2^{\frac12}n^{2n+\frac12}}{e^2\ n^{2n+1}}=\frac{2\sqrt{\pi}}{e^2}\frac{4^n}{\sqrt n}$$

Answer 3

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 $n \geqslant 4$.