Geometric mean of a sequence

$G(m) = \left(\frac{(2m)!}{m!}\right)^{\frac{1}{m}}$

now

$n! \approx \frac{\sqrt{2\pi n}*n^{n}}{e^{n}}$

thus we have the approximation $G(m) \approx \left(\frac{\sqrt{4\pi m}*(2m)^{2m}*e^{m}}{\sqrt{2\pi m}*m^{m}e^{2m}}\right)^{\frac{1}{m}}$

simplifying we get

$\left(\frac{\sqrt{2}*4^{m}m^{m}}{e^{m}}\right)^{\frac{1}{m}}$

$\frac{4m}{e}*\sqrt{2}^{\frac{1}{m}}$

thus

$\frac{G(m)}{m} \approx \frac{4}{e}*\sqrt{2}^{\frac{1}{m}}$

and thus the limit is $\frac{4}{e} \approx \boxed{1.47151776469}$