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Anyway, assuming one can show that $\log \Xi_n = o(\log n)$, presumably via $\Xi_n = o(n)$ almost surely, then it follows (since the $-Z_n^2/2$ term dominates the $-\log Z_n$ term) that:

Here we have to go even further, and assume that $\log \Xi_n = o( \log \log n) ~~ as ~~ n \to \infty$, presumably via $\Xi_n = o(\log n)$, almost surely. Again, all Cramer says is "assuming $\Xi_n$ is bounded". But since all one can say a priori about $\Xi_n$ is that $0 \le Xi_n \le n$ a.s., it hardly seems clear that one should have $\Xi_n = O(1)$ almost surely, which seems to be the substance of Cramer's claim.

Anyway, assuming one can show that $\log \Xi_n = o(\log n)$, then it follows (since the $-Z_n^2/2$ term dominates the $-\log Z_n$ term) that:

Here we have to go even further, and assume that $\log \Xi_n = o( \log \log n) ~~ as ~~ n \to \infty$ almost surely. Again, all Cramer says is "assuming $\Xi_n$ is bounded". But since all one can say a priori about $\Xi_n$ is that $0 \le Xi_n \le n$ a.s., it hardly seems clear that one should have $\Xi_n = O(1)$ almost surely, which seems to be the substance of Cramer's claim.

(Seemingly this would be equivalent to $\log(\Xi_n) = o(\log \log n)$ almost surely via the continuity of $\log/\exp$ and the continuous mapping theorem, but I have not thought about it enough to be sure.)

Thus using the above theorem we have shown that for all $\varepsilon >0$, $\mathbb{P}(Z_n \le u_n^{(\varepsilon)} \text{ i.o.}) = 0$, which to recapitulate should mean that $\Xi_n = o(\log n)$ almost surely.

We need to show still that $\log \Xi_n = o(\log \log n)$. This doesn't follow from the above, since, e.g.,

$$ \frac{1}{n} \log n = o(\log n) \,, - \log n + \log \log n \not= o(\log n) \,. $$

However, given a sequence $x_n$, if one can show that $x_n = o( (\log n)^{\delta})$ for arbitrary $\delta >0$, then it does follow that $\log(x_n) = o(\log \log n)$. Ideally I would like to be able to show this for $\Xi_n$ using the above lemma (assuming it's even true), but am not able to (as of yet).

Anyway the point being that, appealing to this theorem, if we can show that:

$$\sum_{j=1}^{+\infty}[1 - F(u_j^{(\varepsilon)})]\exp(-j[1-F(u_j^{(\varepsilon)})]) = \sum_{j=1}^{+\infty}\left[ \frac{\varepsilon \log j}{j} \right]\exp(-\varepsilon \log j) = \varepsilon \sum_{j=1}^{+\infty} \frac{ \log j}{j^{1 + \varepsilon}} < + \infty \,. $$

Note that since logarithmic growth is slower than any power law growth for any positive power law exponent (logarithms and exponentials are monotonicity preserving, so $\log \log n \le \alpha \log n \iff \log n \le n^{\alpha}$ and the former inequality can always be seen to hold for all $n$ large enough due to the fact that $\log n \le n$ and a change of variables), we have that:

$$ \sum_{j=1}^{+\infty} \frac{\log j}{j^{1 + \varepsilon}} \le \sum_{j=1}^{+\infty} \frac{j^{\varepsilon/2}}{j^{1 + \varepsilon}} = \sum_{j=1}^{+\infty} \frac{1}{j^{1 + \varepsilon/2}} < +\infty \,,$$

since the p-series is known to converge for all $p>1$, and $\varepsilon >0$ of course implies $1 + \varepsilon/2 > 1$.