An Intriguing 'Limits' Problem #2

We know $2^{2n} = \displaystyle \sum_{i=0}^{2n} {2n \choose i}$

We have by A.M - G.M Inequality:

$\dfrac{\displaystyle \sum_{i=0}^{2n} {2n \choose i}}{2n+1} \geq \left(\prod_{i=0}^{2n} {2n \choose i} \right)^{\frac{1}{2n+1}}$

$\Rightarrow \dfrac{2^{2n}}{2n+1} \geq \left( \prod_{i=0}^{2n} {2n \choose i} \right)^{\frac{1}{2n+1}}$

$\Rightarrow \left( \dfrac{2^{2n}}{2n+1} \right)^{\frac{1}{2n}} \geq \left(\prod_{i=0}^{2n} {2n \choose i} \right)^{\frac{1}{2n(2n+1)}}$

$\Rightarrow 2 \geq (2n+1)^{\frac{1}{2n}} \left(\prod_{i=0}^{2n} {2n \choose i} \right)^{\frac{1}{2n(2n+1)}}$

Therefore, when $n$ tends to $\infty$ ,

$\lim_{n \to \infty} (2n+1)^{\frac{1}{2n}} \left(\prod_{i=0}^{2n} {2n \choose i} \right)^{\frac{1}{2n(2n+1)}}= \lim_{n \to \infty} \left( \sqrt{\prod_{i=0}^{2n} {2n \choose i}} \ \right)^{\frac{2n-1}{4n^3 - n}} = \boxed{2}$

nice solution, but i used table on that problem