<s>[INST] For $n \ge 4$, does it follow that ${{3n} \choose {n}} > 4^n$ I believe that the answer is yes.  Here's my thinking:
(1) For $n \ge 4, { {2n} \choose {n}} \ge \frac{4^n}{n}$

By induction:  $64 = \frac{4^4}{4} \le {{8}\choose{4}}= 70$. Assume it is true up to $n-1 \ge 4$.  Then, ${ {2n} \choose n} = 2\left(\frac{2n-1}{n}\right){{2n-1}\choose{n-1}} > 2\left(\frac{2n-1}{n}\right)\frac{4^{n-1}}{n-1} > 2 \cdot 2\cdot \frac{4^{n-1}}{n} = \frac{4^n}{n}$
Note:  This argument was taken from Wikipedia.

(2)  $(3n-1) > \left(\frac{3}{2}\right)(2n-1), (3n-2) > \left(\frac{3}{2}\right)(2n-2), \dots, (2n-n+1) > \left(\frac{3}{2}\right)(2n-n+1)$

If $2a = 3b$, then $2a - 2 > 3b - 3$ and $2(a-1) > 3(b-1)$ so that $(a-1) > \frac{3}{2}(b-1)$

(3)  For $n\ge 2$, ${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n} \choose {n}}$

From (2) above: ${{3n} \choose{n}} = \frac{(3n)(3n-1)\dots(3n-n+2)(3n-n+1)}{n!} > \left(\frac{3}{2}^n\right)\frac{(2n)(2n-1)\dots(2n-n+2)(2n-n+1)}{n!} = \left(\frac{3}{2}^n\right){{2n} \choose {n}}$

(4)  Since for $n \ge 4, \left(\frac{3}{2}\right)^n > n$, it follows that:

${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n}\choose{n}} > n\left(\frac{4^n}{n}\right) = 4^n $
Note: The argument that for $n \ge 2, \left(\frac{3}{2}\right)^n > n$ can be found here.

 [/INST] We have the following Stirling approximations:


*

*$\sqrt{2\pi n}n^ne^{-n} \leq n! \leq e \sqrt{n} n^n e^{-n} $ for all $n \in \mathbb{N}$
$${{3n} \choose {n}} \geq \frac{\sqrt{2\pi 3n}(3n)^{3n}e^{-3n}}{e \sqrt{n} n^n e^{-n} \cdot e \sqrt{2n} (2n)^{2n} e^{-2n}} =\frac{\sqrt{3\pi}}{e^2\sqrt{n}} \frac{27^n}{4^n} > \frac{\sqrt{3\pi}}{e^2} \frac{\left(\frac{3}{2}\right)^n}{\sqrt{n}}4^n$$
So, if $n$ is large enough, the inequality holds.

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