Choosing positive reals $\alpha_i$ for $a_iX_i \sim N(0, a_i^2)$ such that $a_iX_i \stackrel{a.s.}{\to} 0$

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Suppose we have a decreasing sequence $\{\alpha_i\}_{i=1}^{\infty}$ where $\alpha_i \in \mathbb{R}^{+}$. That is

$$\mathbb{R}^{+} \ni \alpha_1 \geq \alpha_2 \geq \cdots > 0.$$

Further, let $\{X_i\}_{i=1}^{\infty} \sim N(0,1)$, and define $Y_i := \alpha_iX_i \sim N(0, \alpha_i^2)$. What conditions does the sequence $\{\alpha_i\}_{i=1}^{\infty}$ need such that $Y_i \stackrel{a.s.}{\to} 0$?

What I have:

1. For each $i$ we want to show that for every $\epsilon_i > 0$ there exists some constant $c_i$ such that

$$\mathbb{P}(|a_iX_i| > c_i) = \mathbb{P}(|Y_i| > c_i) \leq \epsilon_i. \tag{$\star$}$$

So, I need to find some systematic way of choosing $\epsilon_i$ and $\alpha_i$ such that for any $\delta > 0$ we have $\mathbb{P}(|Y_i| > \delta) \to 0$. I've been focusing on a method to incorporate the variances $\alpha_i^2$, which seems important in choosing $c_i$, i.e. the larger the variances, the flatter the pdf of $Y_i$, the larger $c_i$ needs to be to satisfy $(\star)$. I'm not seeing how to easily do this?

2. Once we have (1) satisfied, then we can apply the Borel-Cantelli Lemma to show that for all $\epsilon_i > 0$ we have

$$\sum_{i \geq 1} \mathbb{P}(|Y_i| > \epsilon_i) < \infty \quad\Rightarrow\quad Y_i \stackrel{a.s.}{\to} 0,$$

which in turn answers the question of how to choose sufficient $\{\alpha_i\}_{i=1}^{\infty}$.

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$P(|\alpha_iX_i| >\epsilon)= P(|X_i| >\frac {\epsilon} {\alpha_y})\leq \frac {EX_i^{2}} {(\frac {\epsilon} {\alpha_y})^{2}}=\alpha_i^{2}/\epsilon^{2}$. So $\sum \alpha_i^{2} <\infty$ is a sufficient condition (by Borel Cantelli Lemma).

3
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Hint:

$$ \mathbb{P}[\lvert \alpha_i X_i \rvert > c_i] = \mathbb{P}[\lvert X_i \rvert > \frac{c_i}{\alpha_i}] \leq \frac{\alpha_i}{c_i} \mathbb{E}[\lvert X_i \rvert ] = \frac{\alpha_i}{c_i} \sqrt{\frac{2}{\pi}}$$

by Markovs inequality. For the last inequality I used that the absolute value of a normal distribution follows the half - normal distribution.