Showing that $X_n$ ~ $N(0, a_n)$ converge to $0$ when $a_n \to 0$ sufficiently fast

Question

If $X_n$ have distribution $N(0, a_n)$ with $\sum_{n=1}^\infty a_n^b < \infty$ for some $b > 0$, then $X_n$ converge almost surely to $0$.

I was able to show (for a previous part of the same problem) that $lim_{n \to \infty}a_n = 0$ implies convergence in probability to 0, and I'm reasonably sure that $\sum_{n=1}^\infty a_n^b < \infty$ for some $b > 0$ implies $lim_{n \to \infty}a_n = 0$, so I know this must at least converge in probability (hence, a subsequence converges almost surely) to 0. For almost sure convergence, I would like to make an argument from the Borel-Cantelli lemma, i.e., show that $\sum_{n=1}^\infty P(|X_n| > \epsilon) < \infty$, but when I try to do this by translating the probability into something in terms of the cdf of $|X_n|$ (the half-normal distribution) I consistently get inequalities pointing the wrong way since I'm looking at $1 - F_{|X_n|}(\epsilon)$. Is there a better way to handle this? Note that the $X_n$ are not stated anywhere to be independent.

Answer 1

Fact 1: If $X\sim N(0,a)$ then $\mathbb{P}(|X|>\epsilon)\leq \frac{1}{\epsilon}\sqrt{\frac{2a}{\pi}}\exp[-\epsilon/2a]\leq \exp[-\epsilon/2a]$, whenever $a\leq \epsilon^{2}\pi/2$.

Fact 2: If $t\to 0$ then $t^{-b}e^{-c/t}\to 0$ for any fixed $b,c>0$. In particular $e^{-c/t}<t^{b}$ for sufficiently small $t$.

We know that $a_{n}\to 0$. So there exists $N\in\mathbb{N}$ such that $a_{n}<\epsilon^{2}\pi/2$ and $a_{n}$ is sufficiently small so that we have $e^{-\epsilon/2a_{n}}<a_{n}^{b}$ for all $n>N$. Putting these all together we have

$$\sum_{n=1}^{\infty}\mathbb{P}(|X_{n}|>\epsilon)\leq \frac{1}{\epsilon}\sum_{n=1}^{N}\sqrt{\frac{2a_{n}}{\pi}}\exp[-\epsilon/2a_{n}]+\sum_{n=N+1}^{\infty}a_{n}^{b}<\infty.$$

Therefore by Borel-Cantelli lemma $X_{n}\to 0$ almost surely.