How to prove that $\Bbb P_1(\lim\sup_{n\to\infty}\frac{\log \bf{x}(n)}{\log n}\le \frac{1}{2})=1$, with the suggestion that $\bf{x}(n)$ goes to $\infty$ like $\sqrt{n}$. Here $\bf{x}_n$ is the standard random walk.
I tried to prove this using Borel Cantelli lemma. then we only need to show $\sum_{n=0}^\infty \Bbb P_1(\frac{\log \bf{x}(n)}{\log n} \le \frac{1}{2})=\infty$, which I cannot prove.
Could anyone kindly help? Thank you so much!