When $x$ goes to $+\infty$, the quotient $$\frac{N(x)}{D(x)}=\frac{x+\frac{2}{4!}\,x^4+\frac{2.5}{7!}\,x^7+\frac{2.5.8}{10!}\,x^{10}+\ldots}{1+\frac{1}{3!}\,x^3+\frac{1.4}{6!}\,x^6+\frac{1.4.7}{9!}\,x^9+\ldots}$$ converges rapidly to $$c=\frac{\Gamma(\frac{1}{3})}{\sqrt[3]{3}\Gamma(\frac{2}{3})}\simeq1.37172116$$
As far as I understand, to find the exact value of $c$ requires complex analysis, indefinite integrals, fourier transforms,... But is there a simple proof of the fact that the quotient converges towards a constant, without knowing what it is ? Ideally, I am looking for a proof using the coefficients of the power series in the numerator and denominator.
Furthermore, is there a simple proof of the fact that $cD(x)-N(x)$ is bounded ? That it converges to 0 ?
A straight forward approach would be to first show (in terms of power series) that
$$N(x)=\frac{1}{6}\, 3^{5/6} \;\Gamma \left(\frac{1}{3}\right) \left(\text{Bi}(x)-\sqrt{3}\, \text{Ai}(x)\right)$$
and $$D(x)=\frac{1}{2}\, 3^{1/6} \;\Gamma \left(\frac{2}{3}\right) \left( \text{Bi}(x)+\sqrt{3}\,\text{Ai}(x)\right);$$
$\text{Ai}(x)$ and $\text{Bi}(x)$ being the Airy functions.
Then attempt to find the limit of $$ \frac{\left( \text{Bi}(x)-\sqrt{3}\,\text{Ai}(x)\right)}{\left( \text{Bi}(x)+\sqrt{3}\,\text{Ai}(x)\right)}=\frac{2 \text{Bi}(x)}{\sqrt{3} \text{Ai}(x)+\text{Bi}(x)}-1$$ as $x \rightarrow \infty$; which as it turns out gives a very simple result since $\text{Ai}(x)$ tends to zero as $x \rightarrow \infty$ and $\text{Bi}(x)$ does not.