I have a question (at $\star$) regarding the solution to the question below:
Question. Let $\{X_n\}_{n=1}^{\infty}$ be a sequence of random variables. Show that there exists a sequence of numbers $\{a_n\}_{n=1}^{\infty}$ such that $\frac{X_n}{a_n} \stackrel{a.s.}{\to} 0$.
Solution. For some $b$, note that $\mathbb{P}(|X_n| > b) \to 0$ as $b \to \infty$. So, choose some $b_n > 0$ such that
$$\mathbb{P}(|X_n| > b_n) < \frac{1}{2^n}.$$
Then we can choose $a_n$ as follows:
$$a_n = \max\{n, \max_{k=1, \dots, n} b_k\}.$$
Therefore, for all $\epsilon > 0$ we have
$$\sum_{n=1}^{\infty} \mathbb{P}(|X_n| > \epsilon a_n) \leq \sum_{n=1}^{\infty} \mathbb{P}(|X_n| > a_n) < \infty, \tag{$\star$}$$
and the result follows by Borel-Cantelli.
For $(\star)$, I don't quite see how this holds for all $\epsilon$. Suppose we had a sequence of Gaussians with mean $0$, i.e. let $X_n \sim \mathcal{N}(0, \sigma_n^2)$, and so $|X_n|$ are half-Gaussians. If $\epsilon \geq 1$, then surely $(\star)$ holds, but if $\epsilon < 1$ I don't see how it holds for any sequence of $a_n$.
Where is my flaw in thinking?
Edit: From @Michael's comments, a more elegant (and correct) approach is for each $n \in \mathbb{N}$, define $\epsilon_n := \frac{a_n}{n}$ such that
$$\mathbb{P}(|X_n| > \epsilon_n) < \frac{1}{2^n}.$$
Then
$$\sum_{n=1}^{\infty}\mathbb{P}(|X_n| > \epsilon_n) \leq \sum_{n=1}^{\infty} \frac{1}{2^n} < \infty.$$
Therefore, by B.C. the result follows.