I am attempting Exercise 4.5 from David Williams' Probability with Martingales, which is about Borel-Cantelli lemma. The question states the follows.
If $G$ is a random variable with the normal N(0,1) distribution, then, for $x>0,$ we know that: $$\mathbb{P}(G>x)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{x}e^{-\frac{1}{2}y^2}\,dy\leq\frac{1}{x\sqrt{2\pi}}e^{-\frac{1}{2}x^2}.$$ Then, let $X_1,X_2,\dots$ be a sequence of independent N(0,1) random variables, and define $L:=$limsup$(\frac{X_n}{\sqrt{2\log{n}}}),$ prove that $\mathbb{P}(L\leq1)=1.$
To approach this question and use Borel-Cantelli lemma, I have made some attempts:
$$\mathbb{P}(L\leq1)=\mathbb{P}(\text{limsup}(\frac{X_n}{\sqrt{2\log{n}}})\leq1)\stackrel{*}{=}\mathbb{P}(\frac{X_n}{\sqrt{2\log{n}}}\leq1 \text{ eventually})\\ =\mathbb{P}(X_n\leq\sqrt{2\log{n}} \text{ eventually})=1-\mathbb{P}(X_n>\sqrt{2\log{n}} \text{ infinitly often}).$$ Then, we can use the information given by the question: $$\mathbb{P}(X_n>\sqrt{2\log{n}})\leq \frac{1}{\sqrt{2\log{n}}\sqrt{2\pi}}e^{-\frac{1}{2}\cdot 2\log{n}}=\frac{1}{2n\sqrt{\pi\log{n}}}.$$
But the problem at this step is that $\sum^{\infty}_{n=1}\frac{1}{2n\sqrt{\pi\log{n}}}=\infty,$ and then it seems the Borel-Cantelli lemma can't be applied!
Is there any method to use the Borel-Cantelli lemma to solve this problem? Also, I'm thinking the step of my attempt marked with *: $\mathbb{P}(\text{limsup}(\frac{X_n}{\sqrt{2\log{n}}})\leq1)\stackrel{*}{=}\mathbb{P}(\frac{X_n}{\sqrt{2\log{n}}}\leq1 \text{ eventually}),$ is it technically correct?
Many thanks in advance for any hints and solutions!