Let $\{X_n\}_{n=1}^\infty$ be an i.i.d. sequence of random variables with $\mathbb{E}X_1 = 0, \operatorname{Var}(X_1) = 1$.
Given the fact that $$\limsup_{n \to \infty} \frac{\sum_{i=1}^n X_i}{\sqrt{2n \log \log n}} \le 1 \quad a.s.$$
show that$$\limsup_{n \to \infty} \frac{|\sum_{i=1}^n X_i|}{\sqrt{2n \log \log n}} \leq 1 \quad a.s.$$
My attempt:
Denote the probability space on which we are working with $\Omega$. Assume that $$\limsup_{n \to \infty} \frac{\sum_{i=1}^n X_i(\omega)}{\sqrt{2n \log \log n}} \leq 1$$
Then, choosing a subsequence $(n_k)_{k=1}^\infty$ such that
$$\limsup_{n \to \infty} \frac{|\sum_{i=1}^n X_i(\omega)|}{\sqrt{2n \log \log n}} = \lim_{k \to \infty} \frac{|\sum_{i=1}^{n_k} X_i(\omega)|}{\sqrt{2 n_k \log \log n_k}}$$ we get $$\limsup_{n \to \infty} \frac{|\sum_{i=1}^n X_i(\omega)|}{\sqrt{2n \log \log n}} = \left| \lim_{k \to \infty} \frac{\sum_{i=1}^{n_k} X_i(\omega)}{\sqrt{2 n_k \log \log n_k}}\right|$$
and if I could argue that this last limit is positive (almost everywhere), then I can conclude what I want.
How do I proceed?
You have to apply the hypothesis to the sequence $(-X_n)$ which also has mean $0$ and variance $1$. Then use the fact that $\lim \sup a_n \leq 1$ and $\lim \sup (-a_n) \leq 1$ implies $\lim \sup |a_n| \leq 1$ for any sequence of real numbers $(a_n)$.
If possible let $\lim \sup |a_n| >1$. Let $r$ be such that $\lim \sup |a_n| >r>1$. Then $|a_{n_k}| \geq r$ for some subequence ${n_k}$. Now either $a_{n_k} \geq 0$ for infinitely many $k$ or $a_{n_k} \leq 0$ for infinitely many $k$. Hence $a_n \geq r$ for infinitely many $n$ or $-a_n \geq r$ for infinitely many $n$. But then $\lim \sup a_n \geq r >1$ or $\lim \sup (-a_n) \geq r>1$ leading to a contradiction.