Interesting Limit (21)

We have the formula for $a,b\in\mathbb N$ and setting $a=b=1$ we deduce $(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r} b^r= \sum_{r=0}^{n}\binom{n}{r}=2^n$

and hence our $\sqrt[n]A_n=\dfrac{2}{\sqrt[n]n}$ . Let us denote $X_n= \prod_{r=1}^{n-1}\binom{n}{r}=\dfrac{1}{(n!)^{n+1}}\left(\prod_{r=1}^n r^r\right)^2$ Also we have the approximation for $\dfrac{1}{(n!)^{n+1}}\left(\prod_{r=1}^{n} r^r\right)^2 \\ \approx\dfrac{1}{(n!)^{n+1}}\left( A n^{\frac{6n^2+6n+1}{12}}e^{-\frac{n^2}{4}}\right)^2$ using the Stirling approximation for $n!$ and simplifying we have $X_n\approx \dfrac{A^2e^{\frac{n^2+2n}{2}}}{(2\pi)^{\frac{n+1}{2}}n^{\frac{3n+2}{6}}}$ where $A$ is Glashier-Kinkelin Constant and required we have $\sqrt[n]G_n =\left(\sqrt[n]{X_n} \right)^{\frac{1}{n}}\approx \dfrac{A^{\frac{2}{n^2}}e^{\frac{1+n}{2n}}}{\sqrt[2n]{2\pi n}}$ Therefore our limit $\lim_{n\to \infty}\left(\dfrac{G_n}{A_n}\right)^{\frac{1}{n}}=\lim_{n\to \infty}\dfrac{\sqrt[n] n }{2}\cdot \dfrac{A^{\frac{2}{n^2}}e^{\frac{1+n}{2n}}}{\sqrt[2n]{2\pi n}}=\dfrac{e^{\frac{1}{2}}}{2}$

Naren is correct