Limits with AlphaBets

Note that the numerator is nothing but $\displaystyle \left(G(n+2)\right) ^{\frac {1}{n^2}}$

Where $G(z)$ is the Barnes G-function. Take log with base $e$ sides to get $\displaystyle \lim_{n\to \infty} \left(\frac {1}{n^2}\ln (G(n+2)) -\alpha \ln n\right)= \beta$

Using the asymptotic expansion of Barnes G-function and adjusting the coefficients(for limit existence) we get $\alpha=\frac 12$ and $\beta =-\frac 34$

PS:

Asymptotic expansion of Barnes G-function is $\displaystyle \ln(G(n+1)) \sim\frac {n^2}{2}\ln n -\frac {1}{12}\ln n -\frac {9n^2 -1}{12} +\frac n2 \ln 2\pi -\ln A$

Where $A$ is Glaisher-Kinkelin constant