Question: Show that if $X_k$, $k\in\Bbb N$ are iid standard normal variables, then there exists a constant $C$ such that for sufficiently large $k\ge2$ we have $|X_k|\le C\sqrt{\log k}$ almost surely.
To me this isn't even true at first glance because the law of $X_k$ doesn't depend on $k$ and for any $r\in\Bbb R$ (taking $r$ as an arbitrary bound, in particular it can be equal to $C\sqrt{\log k}$) $\Bbb P(|X|> r)>0$ so $\Bbb P(|X|\le r)<1$ not matter how big we choose $C$ or $k$.
Am I wrong or is the question wrong?
Since the random variables are identically distributed, we have by Markov's inequality
\begin{align*} \mathbb{P}(|X_n| \geq 6 \sqrt{\log n})=\mathbb{P}(|X_1| \geq 6 \sqrt{\log n}) &= \mathbb{P}\left( \frac{X_1^2}{4} \geq 9 \log n \right) \\ &= \mathbb{P}(e^{X_1^2/4} \geq e^{9 \log n}) \\ &\leq \frac{1}{e^{\log n^9}} \mathbb{E}\exp(X_1^2/4) \\ &= \frac{1}{n^9} \underbrace{\mathbb{E}\exp(X_1^2/4)}_{<\infty}.\end{align*}
(Note that $\mathbb{E}\exp(X_1^2/4)$ is finite because $X_1$ is standard Gaussian.) In particular,
$$\sum_{n \geq 1} \mathbb{P}(|X_n| \geq 6 \sqrt{\log n}) < \infty.$$
By the Borel-Cantelli lemma, this implies that $|X_n| \leq 6 \sqrt{\log n}$ with probability $1$ for $n=n(\omega)$ sufficiently large.
Remark: With some more effort, one can show a stronger result.