Let $G_1, \dots, G_n$ be iid random variables, such that $G_1 \sim \mathcal N(0,1)$

Let $$S_n = \sum_{i=1}^n G_i\quad \text{and} \quad\tilde{S}_n = \frac{1}{\sqrt{2n\log\log n}}S_n$$

It is easy to see that $\tilde{S}_n \sim N(0,\sigma_n)$, where $\sigma_n = \frac{1}{\sqrt{2\log\log n}}$.

By the iterated logarithm law, $$\limsup_{n \rightarrow \infty} \ \tilde{S}_n = 1 \quad \text{a.s} $$

So what is troubling for me, is that I cannot prove by direct means that if $(Y_n)_{n \in \mathbb N}$ are indepedent random variables such that $Y_n \sim \mathcal N(0,\sigma_n)$ then it holds that $$\limsup_{n \rightarrow \infty} \ Y_n = 1 \quad \text{a.s} $$

For instance, only trying to prove that : a.s, $Y_n > 1+\epsilon$ occurs only a finite number of times :

Using Borell-Cantelli we want to prove that $$\sum_{n} \mathbb P(Y_n>1+\epsilon) <\infty $$ However, Gaussian concentration only gives : $$\mathbb P(Y_n>x) = P(\sigma_nG>x) \leq e^{-\frac{x^2}{2\sigma_n^2}} $$

One can check it is far from summable. So where do we loose information ? Is the inequality too loose ?