$\lim_m \mathbb{P}\left\{ \left|S_n\right| > b \sqrt{n} \text{ for some }n\le m\right\} = 1\,?$ Let $b$ be some positive number (say $b = 1.96$). Let $X_1,\cdots,X_n$ be a sequence of iid random variables such that they are mean zero and variance $1$. Let $S_n = X_1 + \cdots + X_n$. How can I show that
$$\lim_m \mathbb{P}\left\{ \left|S_n\right| > b \sqrt{n} \text{ for some }n\le m\right\} = 1\,?$$
The hint says law of iterated logarithm, which states that
$$\mathbb{P}\left\{ \limsup_{n\to\infty} \frac{S_n}{\sqrt{2n \log\log n}} =1\right\}=1.$$
I have some hand-wavy thoughts (in fact I guess I'm close), but not sure how to write a concrete proof. LIL suggests that for almost all $\omega\in\Omega$ and any $\epsilon>0$,
$$\left|\sup_{n\ge m} \frac{S_n}{\sqrt{2n\log \log n}}-1\right| < \epsilon$$
for large $m$. This implies
$$1-\epsilon< \sup_{n\ge m}\frac{S_n}{\sqrt{2n \log \log n}}<1+\epsilon$$
Which implies
$$\frac{S_n}{\sqrt{2n \log \log n}}< 1+\epsilon \,\,\forall n\ge m \text{, and }1-\epsilon < \frac{S_k}{\sqrt{2k \log \log k}} \text{ for some }k\ge m$$
The second one implies
$$\frac{\left|S_k\right|}{\sqrt{k}}>\left(1-\epsilon\right)\sqrt{2 \log \log k}>\left(1-\epsilon\right) \sqrt{2 \log \log m}$$
But how should I proceed? First, $m$ depends on $\epsilon$, so I'm not sure if I can make $\left(1-\epsilon\right)\sqrt{2 \log \log m} \ge b$. But second, even if I can, this would only show that
$$\mathbb{P}\left\{\text{Given }b,\text{ there exists }k \text{ s.t. }\left|S_k\right| > b\sqrt{k} \right\} = 1$$
Update: So after some thoughts I think I came up with the following proof. Still, any alternative/simpler proof is welcomes!
Almost surely, we have
$$\forall \epsilon>0, \exists M_\epsilon \text{ s.t. } \left| \sup_{n\ge m} \frac{S_n}{\sqrt{2n\log \log n}} - 1 \right|< \epsilon \text{ for all } m \ge M_\epsilon$$
In particular, almost surely, we have
$$\left|\sup_{n\ge m} \frac{S_n}{\sqrt{2 n \log \log n}} - 1 \right| < \frac{1}{2} \text{ for all }m\ge M_{1/2}$$
Given $b > 0$, there exists $N$ such that $\frac{1}{2} \sqrt{2 \log \log N}= b$. Then, almost surely, we have
$$\left|\sup_{n\ge m} \frac{S_n}{\sqrt{2 n \log \log n}} - 1 \right| < \frac{1}{2} \text{ for all }m\ge \max\left(M_{1/2},N\right)$$
This implies that almost surely we have
$$\frac{S_k}{\sqrt{2k \log \log k}} >\frac{1}{2}\text{ for some }k\ge \max\left(M_{1/2},N\right)$$
In particular, this implies
$$\mathbb{P}\left\{ \frac{\left|S_k\right|}{\sqrt{k}} >  b \text{ for some }k\ge \max\left(M_{1/2}, N\right)\right\} = 1$$
 A: You want to prove that $$\lim_{m\rightarrow \infty}\mathbb P\left(\bigcup_{n \le m} \left\{\frac{|S_n|}{\sqrt n} > b\right\}\right) = 1$$ Or $$\mathbb P\left(\bigcap_{n \ge 0} \left\{\frac{|S_n|}{\sqrt n} \le b\right\}\right) = \lim_{m\rightarrow \infty}\mathbb P\left(\bigcap_{n \le m} \left\{\frac{|S_n|}{\sqrt n} \le b\right\}\right) = 0$$
First you can observe that $\mathbb{P}\left\{ \limsup_{n\to\infty} \frac{S_n}{\sqrt{2n \log\log n}} =1\right\}=1$ implies that there is $\mathcal N$, such that $\mathbb P (\mathcal N) = 0$ and $$\forall \omega \in \Omega \backslash \mathcal N,~ \forall p \in \mathbb N, ~\exists n\ge p,~ \left| \frac{S_n}{\sqrt{2n \log\log n}}-1\right| < \frac12 $$ which can be written $$\forall p \in \mathbb N,~ \mathbb P\left(\bigcap_{n\ge p} \left\{\left| \frac{S_n}{\sqrt{2n \log\log n}}-1\right| \ge \frac12\right\}\right) = 0$$ Now let $N\in \mathbb N$ such that $b \le \frac12\sqrt {2 \log \log N}$. As $$\left\{\frac {|S_n|}{\sqrt n} \le b\right\} \subset \left\{-\frac{b}{\sqrt{2\log\log n}} - 1 \le \frac{S_n}{\sqrt{2n \log\log n}}-1 \le \frac{b}{\sqrt{2\log\log n}} - 1\right\} \subset \left\{ \left| \frac{S_n}{\sqrt{2n \log\log n}}-1\right| \ge \frac12 \right\},~~ \forall n \ge N $$ Now $$\bigcap_{n \ge 0} \left\{\frac{|S_n|}{\sqrt n} \le b\right\} \subset \bigcap_{n \ge N} \left\{\frac{|S_n|}{\sqrt n} \le b\right\} \subset \bigcap_{n \ge N} \left\{\left| \frac{S_n}{\sqrt{2n \log\log n}}-1\right| \ge \frac12\right\}$$ and you have your answer after that.
