Determine or find an upper bound for
$$\limsup_{n \to \infty} \left|\frac{\frac{2}{n} (l(\beta) - l(\hat{\beta})) - (\beta - \hat{\beta})^T \nabla^2 l(\hat{\beta})(\beta - \hat{\beta})}{(\beta - \hat{\beta})^T \nabla^2 l(\hat{\beta}) (\beta - \hat{\beta})}\right|$$
where $l$ is the log-likelihood function in a model of logistic regression with $n$ samples of the form $(y_i,X_i)$, where $y_i \in \{0,1\}$ and $X_i \in R^p$, $\beta \in R^p$ is an unknown vector of parameters and $\hat{\beta}$ is the MLE.
W.l.o.g. I assume that $y_i = 0$ for all $i = 1,\dots,n$ and rearrange the expression to
$$\left|\limsup_{n \to \infty} \frac{\frac{2}{n} \sum_{i=1}^n \text{log}\left(\frac{1+e^{X_i^T\beta}}{1+e^{X_i^T\hat{\beta}}}\right) - (\beta - \hat{\beta})^T X^T \text{diag}\left(\frac{e^{X_1\hat{\beta}}}{(1+e^{X_1\hat{\beta}})^2},\dots,\frac{e^{X_n\hat{\beta}}}{\left(1+e^{X_n\hat{\beta}}\right)^2}\right)X(\beta - \hat{\beta})}{(\beta - \hat{\beta})^T X^T \text{diag}\left(\frac{e^{X_1\hat{\beta}}}{\left(1+e^{X_1\hat{\beta}}\right)^2},\dots,\frac{e^{X_n\hat{\beta}}}{\left(1+e^{X_n\hat{\beta}}\right)^2}\right)X(\beta - \hat{\beta})}\right|$$
$$= \left|\limsup_{n \to \infty} \frac{\frac{2}{n}\sum_{i=1}^n \text{log}\left(\frac{1+e^{\sum_{j=1}^p x_{ij}\beta_j}}{1+e^{\sum_{j=1}^p x_{ij}\hat{\beta}_j}}\right) - \sum_{s=1}^p \sum_{m=1}^p \sum_{i=1}^n \frac{e^{\sum_{j=1}^p x_{ij}\beta_j}}{\left(1+e^{\sum_{j=1}^p x_{ij}\hat{\beta_j}}\right)^2}x_{im}x_{is}(\beta_m - \hat{\beta}_m)(\beta_s - \hat{\beta}_s)}{\sum_{s=1}^p \sum_{m=1}^p \sum_{i=1}^n \frac{e^{\sum_{j=1}^p x_{ij}\beta_j}}{\left(1+e^{\sum_{j=1}^p x_{ij}\hat{\beta_j}}\right)^2}x_{im}x_{is}(\beta_m - \hat{\beta}_m)(\beta_s - \hat{\beta}_s)}\right|$$
An idea would be to use an application of the Borel-Cantelli lemma
Let $a \in R$ be such that
$$ \sum_{n=1}^{\infty} P(X_n > a) < \infty.$$
Then
$$\limsup_{n \to \infty} X_n \leq a$$
almost surely.
For my purposes it would be awesome if I could show this either for $a = 0$ or for $a = \frac{l(\beta)-l(\hat{\beta}) - \frac{1}{2} (\beta -\hat{\beta})^T \nabla^2 l(\hat{\beta}) (\beta - \hat{\beta})}{\lVert\beta - \hat{\beta}\rVert^2}$.
Another idea would be to use the law of large numbers by showing that
$$\mathbb{E}\left[\frac{1}{n} \sum_{i=1}^n \text{log}\left(\frac{1+e^{X_i^T\beta}}{1+e^{X_i^T\hat{\beta}}}\right)\right] = \frac{1}{2}(\beta - \hat{\beta})^T X^T \text{diag}\left(\frac{e^{X_1\hat{\beta}}}{(1+e^{X_1\hat{\beta}})^2},\dots,\frac{e^{X_n\hat{\beta}}}{\left(1+e^{X_n\hat{\beta}}\right)^2}\right)X(\beta - \hat{\beta})$$
but I wouldn´t know how, especially since the distribution of $X_i$ is not specified.
Both ideas would only deliver almost sure statements though but that would be better than nothing.