Question
Let $(a_n) \subseteq \mathbb R_+$ be a sequence of non-negative numbers such that $$ \limsup_{n \to \infty} \left[ \frac{a_n}{n^\varepsilon} \right]^{1/\log \log n} = e^{1+\varepsilon} $$ for some $\varepsilon > 0$.
Why does it follow that $a_n = O(n^\varepsilon \log^{1+\varepsilon} n)$ ?
Thoughts/Attempt
By definition of big-O notation and the non-negativity of $(a_n)$, we need to show $\limsup_{n \to \infty} \frac{a_n}{n^\varepsilon \log^{1+\varepsilon} n} < \infty$.
I'm trying to manipulate $\frac{a_n}{n^\varepsilon \log^{1+\varepsilon} n}$ to make use of the given limes superior. For example, $$ b_n := \frac{a_n}{n^\varepsilon \log^{1+\varepsilon} n} = \frac{ \left( \left[ \frac{a_n}{n^\varepsilon} \right]^{1/\log \log n} \right)^{\log \log n} }{ \log^{1+\varepsilon} n}. $$ Hence, $$ \begin{aligned} \log b_n &= \log \log (n) \cdot \log\left[ \Big( \frac{a_n}{n^\varepsilon} \Big)^{1/\log \log n} \right] - (1+\varepsilon) \log \log (n) \\ &= \log \log (n) \cdot \Big[ \log\left[ \Big( \frac{a_n}{n^\varepsilon} \Big)^{1/\log \log n} \right] - (1+ \varepsilon)\Big]. \end{aligned} $$ The limes superior of the factor in the square brackets converges to $0$, since $$ \begin{aligned} \limsup_{n \to \infty} \Big[ \log\left[ \Big( \frac{a_n}{n^\varepsilon} \Big)^{1/\log \log n} \right] - (1+ \varepsilon)\Big] &= \log\left[ \limsup_{n \to \infty} \Big( \frac{a_n}{n^\varepsilon} \Big)^{1/\log \log n} \right] - (1+ \varepsilon) \\ &= \log(e^{1+\varepsilon}) - (1+\varepsilon) = 0. \end{aligned} $$ But $\log \log n \to \infty$, so it's not clear that $\limsup_{n \to \infty} \log b_n = \log(\limsup_{n \to \infty} b_n) < \infty$.
Where this comes from
I'm reading the paper "Testing for (in)finite moments" by Lorenzo Trapani, in particular the proof of Theorem 1.
There's a lemma showing that $$ \limsup_{n \to \infty} \left[ \frac{\hat \mu_n }{n^{\frac{k}{\gamma} - 1}} \right]^{1/\log \log n} = e^{k/\gamma} \quad \text{almost surely,} $$ where $(\hat \mu_n)$ is a sequence of non-negative random variables and $0 < \gamma < k$. (Note: This is a slight modification of the paper's notation. There it's called $\hat \mu_k$, but is in fact a sequence in $n$, where the the dependence on $n$ has been suppressed.)
The author states that this implies that $\hat \mu_n = O_{\text{a.s.}}\left( n^{\frac{k}{\gamma}-1} \log^{k / \gamma} n \right)$.
It is not true. For $a_n = n^\varepsilon \log ^{1 + \varepsilon } (n)\log \log (n)$ (with $n\geq 3$) it holds that $$ \mathop {\lim \sup }\limits_{n \to + \infty } \left[ {\frac{{a_n }}{{n^\varepsilon }}} \right]^{1/\log \log (n)} = \mathop {\lim }\limits_{n \to + \infty } \left[ {\frac{{a_n }}{{n^\varepsilon }}} \right]^{1/\log \log (n)} = e^{1 + \varepsilon } , $$ yet $a_n \neq \mathcal{O}(n^\varepsilon \log ^{1 + \varepsilon } (n))$.