Prove that the sequence $X_n = Z_n/ \sqrt{1+\log{n}}, ~ n\geq 1$ is almost surely bounded for $Z_n ~ i.i.d. ~ N(0,1)$. Let $Z_1,Z_2,\ldots$ are i.i.d. $N(0,1)$ random variables. Let $X_n = Z_n/ \sqrt{1+\log{n}}, ~ n\geq 1$. Question is to show that $X_n$ is almost surely bounded. 
Proceeding as following :
Step 1. I have a proof that $P(|X_n|\geq \lambda)\leq n^{-\lambda^2/2} \exp(-\lambda^2/2)$.
Step 2. Then how do I bound $P(sup_{n\geq 1} |X_n| > \lambda)$ for $\lambda \geq 2$, so that I can show that $X_n$ is almost surely bounded?
 A: Hint: $\sup |X_n|$ is greater than $\lambda$ if and only if at least one of the $|X_n|>\lambda$. So you need a bound for the probability of the union:
$$
\mathbb P\left(\bigcup_{n\ge0} |X_n|>\lambda\right)\le \dots
$$
What ways do you know to bound the probability of the union of events?
A: In contrast to how Mike Earnest has suggested (although I'm sure his suggestion will work!), you can use a Borel-Cantelli argument. You say you have the bound
$$ P(|X_n| \ge \lambda) \le n^{-\lambda^2/2} \exp(-\lambda^2/2). \tag{1} $$
Now let us define events
$$ A_n = \{ |X_n| \ge \alpha \}. $$
By $(1)$, we have that
$$ P(A_n) \le n^{-\alpha^2/2}. $$
For Borel-Cantelli, we want $\sum_n P(A_n)$ to converge, ie for the infinite sum to exist (and be finite). This says that we may take, for example, $\alpha = 2$, as then $P(A_n) \le n^{-2}$ and so $\sum_n P(A_n) < \infty$ (the exact value is irrelevant; it just matters that it's finite). Hence Borel-Cantelli then says that
$$ P(|X_n| \ge 2 \text{ i.o.} ) = 0, 
\quad \text{so} \quad
P({\limsup}_n |X_n| \le 2) = 1.$$
Once you have that the $\limsup$ is bounded, you can transfer this over to the whole process; I leave this as an exercise -- which you have completed correctly in the comment below.
Solution by the OP to the exercise (included a suggestion from Sam) :
Fix $\epsilon > 0$. $P(\lim \sup_n |X_n| \leq 2) = 1 \implies \lim_m P(\cap_{n \geq m} \{X_n \leq 2\}) =1 \implies$ $$\exists N\in \mathbb{N} \text{ s.t. } P(\cap_{n \geq N} \{X_n \leq 2\}) \geq 1-\epsilon /2$$
Noting $X_i$ i.i.d. $N(0,1)$, $\exists M>2$ s.t. $P(|X_n| > M) < \epsilon/(2M)$. So, $$P(\cup_{n<N} |X_n| > M) \leq \sum_{n<N} P(|X_n|>M) < \epsilon/2$$
Now, $$P(\cap_{n\in \mathbb{N}} \{X_n \leq M\}) \geq P(\cap_{n\geq N} \{X_n \leq M\}) - P(\cup_{n<N} |X_n| > M) > 1-\epsilon$$
Since $\epsilon >0$ arbitrary we have that $P(\cap_{n\in \mathbb{N}}\{X_n \leq M\}) = 1 \implies P(\sup |X_n| \leq M) = 1$.

