Show that almost surely $\limsup_{n \rightarrow \infty} \frac{\log Z_n}{\log \mathbb{E}[Z_n]} \leq 1$ for any positive, non-decreasing random variables $Z_n$ such that $Z_n \rightarrow \infty$ a.s. as $n \rightarrow \infty$.
I have attempted to use the Etemadi's proof technique for the strong law of large numbers, but the choice of subsequence $n_k$ is dependent on the growth rate of $\log \mathbb{E}[Z_n]$ which we have no control over. So I'm not sure how best to pick a subsequence. (In the SLLN, we divide by $n$ which grows polynomially in $n$ and hence $n_\ell = \alpha^\ell$ works for any $\alpha > 1$ in that proof. But again, there the growth rate is well-understood). My best guess is that a growth rate must be picked which allows to apply Borel-Cantelli on a subsequence, so I thought taking $n_k = \inf\{n: \log\mathbb{E}[Z_n] \leq k^2\}$ might work, but figuring out the details has been a bit tricky.
The hypothesis that $0 \le Z_n \to \infty$ a.s. can be replaced by the weaker hypothesis that $ Z_n \ge 0 $ and $E(Z_n) \to \infty$. The hypothesis that $Z_n$ are nondecreasing will be used in $(2)$ below.
Fix $r$ such that $E(Z_r) >1$. For each $n \ge r$, there is a unique $k_n \ge 1$ such that $$(k_n-1)^2 <\log E(Z_n) \le k_n^2 \,.$$
For every $k \ge k_r$, let $\ell(k)$ be the largest $\ell$ so that $$E(Z_\ell) \le e^{k^2} \,.$$
Markov's inequality gives $$P\Bigl(Z_{\ell(k)} \ge e^{k+k^2} \Bigr) \le \frac {E(Z_{\ell(k)})}{e^{k+k^2}} \le e^{-k} \,, $$ so by Borel-Cantelli, with probability 1 there exists some (random) $k_*$ such that $$Z_{\ell(k)} < e^{k+k^2} \; \, \, \text{for all} \; \, k \ge k_* \,. \tag{1}$$
For all $n \ge r$ we have $n \le \ell(k_n)$. Thus if $k_n \ge k_*$, then $$\frac{\log Z_n}{\log E(Z_n)} \le \frac{\log Z_{\ell(k_n)}} {(k_n-1)^2} \le \frac{k_n+ k_n^2}{(k_n-1)^2} \,, \tag{2}$$ so $$\limsup_{n \to \infty} \frac{\log Z_n}{\log E(Z_n)} \le \limsup_{n \to \infty} \frac{k_n+ k_n^2}{(k_n-1)^2} =1 \,.$$