Let $(X_n)_{n \geq 1}$ be a family of independent, identically distributed integer valued random variables with $\mathbb{P}(X_n \geq i) = 1/i$ for each $i \geq 1$. For each $n$, define $M_n = \max\{X_k \mid 1 \leq k \leq n\}$. Then I am trying to show that $$ \frac{\log M_n}{\log n} \overset{a.s.}{\longrightarrow} 1 \quad \text{as } n \to \infty $$
My attempt
Using the first and second Borel-Cantelli lemmas, it is not difficult to show that for $\alpha > 0$, we have $$ \mathbb{P}(X_n \geq n^\alpha \text{ infinitely often}) = \begin{cases} 0 \quad \alpha > 1 \\ 1 \quad \alpha \leq 1 \end{cases} $$
We also have
$$
\mathbb{P}\Big(\limsup_n \frac{\log{M_n}}{\log n} \geq \alpha\Big) =\mathbb{P}\Big(\frac{\log{X_n}}{\log{n}} \geq \alpha \text{ i.o.}\Big) = \mathbb{P}(X_n \geq n^\alpha \text{ i.o.})
$$
From the two results above, we see that $$ \limsup_n \frac{\log{M_n}}{\log n} \leq 1 $$ almost surely. Now, it suffices to show that the $\liminf$ of the same expression is greater than or equal to $1$. This direction causes me more difficulty.
Note that, for $\epsilon > 0$, we have \begin{align} \mathbb{P}\Big(\liminf_n\frac{\log M_n}{\log n} < 1 - \epsilon\Big) &= \mathbb{P}\Big(\frac{\log M_n}{\log n} <1 - \epsilon \text{ i.o.}\Big) \\ &= \mathbb{P}\Big(M_n <n^{1-\epsilon} \text{ i.o.}\Big) \end{align}
And also
\begin{align} \mathbb{P}(M_n <n^{1-\epsilon}) &= \mathbb{P}(X_k <n^{1-\epsilon} \text{ for } k <n) \\ &= \prod_{k=1}^n\mathbb{P}(X_k <n^{1-\epsilon}) \\ &= \mathbb{P}(X_1 <n^{1-\epsilon})^n \end{align}
We then have
\begin{align} \mathbb{P}(X_1 < n^{1-\epsilon}) &= 1 - \mathbb{P}(X_1 \geq n^{1-\epsilon}) \\ &=1- \mathbb{P}(X_1 \geq \lceil n^{1-\epsilon} \rceil) \\ &= 1 - \frac{1}{\lceil n^{1-\epsilon} \rceil} \end{align}
My problem is this ceiling function. Suppose that instead we had $$ \mathbb{P}(X_1 < n^{1-\epsilon}) = 1 - \frac{1}{n^{1-\epsilon}} $$
Then we would have
\begin{align} \mathbb{P}(X_1 <n^{1-\epsilon})^n &= \Big(1 - \frac{n^{\epsilon}}{n}\Big)^n \\ &\leq e^{-n^{\epsilon}} \end{align}
Then since $$ \sum_{n=0}^\infty e^{-n^\epsilon} < \infty $$ the result would follow by the second Borel-Cantelli lemma. However, I'm not sure how to adapt the argument in light of the ceiling function. It seems clear that the same sort of reasoning will work, because the ceiling function is "very close" to the actual value, but I don't know how to make that rigorous.