to section 5 of @imj. We recall his notations, slightly modified to avoid confusions: $N_+=N+1/2$, $n_+=n+1/2$, $z=\mu_N+\sigma_N s$, with $\mu_N=(\sqrt{(N+\alpha)_+}+\sqrt{N_+})^{1/2}$ and $\sigma_N=(\sqrt{(N+\alpha)_+}+\sqrt{N_+})(1/\sqrt{N_+}+1/\sqrt{(N+\alpha)_+})^{1/3}$. For reasons that will be transparent later on, our aim is to show that $$\label{pointwisecv}
F_N(z)=(-1)^N \sigma_N^{-1/2}\sqrt{N!/n!}\,z^{(\alpha+1)/2}{\mathrm{e}}^{-z/2}L_N^{\alpha_N}(z) {\rightarrow}{\mathrm{Ai}}(s), \;
\forall s \in \mathbb{R} \; ,$$ and $$\label{uniformcv}
F_N(z)={\mathrm{O}}({\mathrm{e}}^{-s})\; \text{uniformly in } \; [s_0,\infty), s_0 \in \mathbb{R} \; .$$

The scaling is slightly different from the original proof: $N^{-1/6}$ has been replaced by ${\sigma_N}^{-1/2}$. As in @imj, we focus on $w_N(z)=z^{(\alpha+1)/2}{\mathrm{e}}^{-z/2}L_N^{\alpha}(z)$, which satisfies $$\frac{d^2w}{dz^2}=\left(\frac{1}{4}-\frac{\kappa}{z}+\frac{\lambda^2-1/4}{z^2}\right)w \; ,$$ where $\kappa=N+(\alpha+1)/2$ and $\lambda=\alpha/2$. Remark that under ${\mathbf{AB}}{\overset{\text{def}}{\Leftrightarrow}}n,N,n/N {\rightarrow}\infty$, $\kappa \sim \lambda$. Our strategy is to reformulate the problem in terms of so-called Whittaker functions, denoted $W_{k,m}$, and to use the extensive available studies of these functions to show (\[whattoshow1\]) and (\[whattoshow2\]). @temme, formula (3.1) p.117 shows that $$w_N(z)=\frac{(-1)^N}{N!}W_{\kappa,\lambda}(z) \; .$$

From now on, we will closely follow @olver80. Let us remark that $$F_N(z)={\sigma_N}^{-1/2}\frac{1}{\sqrt{n!N!}}{W_{\kappa,\lambda}}(z) \;.$$ We fix $s_0 \in \mathbb{R}$, and we work only with $z=\mu_N+{\sigma_N}s$, where $s\geq s_0$.

#### Preliminaries

Following @olver80, we introduce $l=\kappa/\lambda$, $\beta=\sqrt{2(l-1)}$, and the turning points $x_1=2l-2\sqrt{l^2-1}$, $x_2=2l+2\sqrt{l^2-1}$, after the rescaling $x=z/\lambda$. We remark that the two turning points coalesce at 2 under the hypothesis ${\mathbf{AB}}$. In the 