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The exact question I have is $$\lim_{n\to\infty} \frac{G(n+\frac{5}{4})G(n+\frac{7}{4})^{2}G(n+\frac{9}{4})}{G(n+2)^{2}G(n+\frac{3}{2})^{2}}$$

Intuitively it looks like it should tend to $1$ but I do not have experience with the Barnes G-function.

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  • $\begingroup$ Does $G(n+2)^2$ mean $G(n+2) \cdot G(n+2)$ or $G(n^2+4n+4)$? $\endgroup$ – TheSimpliFire Dec 22 '17 at 19:40
  • $\begingroup$ $[G(n+1)]^{2}$ apologies for the confusion $\endgroup$ – Joshua Farrell Dec 22 '17 at 19:48
  • $\begingroup$ You could take logarithms, and use the asymptotics of $\log G(z+1)$ found here, I'd imagine $N=1$ is sufficient. No, it's not pretty, but who promised it would be? BTW, I'd be very much surprised if the limit were $1$. $\endgroup$ – Professor Vector Dec 22 '17 at 20:31
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Use the known asymptotics $$ \ln G(1+z)\sim z^2\left(\frac{1}{2}\ln z-\frac{3}{4}\right)+\frac{1}{2}\ln (2\pi)z-\frac{1}{12}\ln z+\ldots $$ to deal with your limit $\lim_{n\to\infty}f(n)$. Consider $\ln f(n)$: $$ \ln f(n)=\ln G((n+1/4)+1)+2\ln G((n+3/4)+1)+\ln G((n+5/4)+1)-2\ln G((n+1)+1)-2\ln G((n+1/2)+1) $$ $$ \sim \frac{1}{8}\ln n\ , $$ from which one deduce that $f(n)\sim n^{1/8}\to \infty$.

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In the same spirit as Pierpaolo Vivo's answer, using the given asymptotics with more terms, you can have a quite accurate approximation of $$f_n=\frac{G(n+\frac{5}{4})\,G(n+\frac{7}{4})^{2}\,G(n+\frac{9}{4})}{G(n+2)^{2}\,G(n+\frac{3}{2})^{2}}$$ even for small values of $n$.

Expanding $\log(f_n)$ and using later $f_n=e^{\log(f_n)}$, we should get $$f_n= {n^{1/8}}\left(1+\frac{3}{32 n}-\frac{51}{2048 n^2}+\frac{405}{65536 n^3}+O\left(\frac{1}{n^4}\right) \right)$$ which, for sure, shows the limit.

Concerning the values for small $n$, the approximation gives $10$ exact significant figures as soon as $n > 22$.

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