The formal question is as follow: Let $X_1,...,X_N,...$ are independent random variables with distribution $\text{Pr}(X_i=1)=1/2$ and $\text{Pr}(X_i=-1)=1/2$, denote $$Y_N=\text{Max}_{1\leq i\leq N}|\sum_{j=1}^{i}X_j|$$ which is the maximal absolute value achieved in a random process.
I want to estimate the limit of distributions of $\frac{Y_N}{N}$ when $N$ goes to infinity. Especially, I want to get a upper bound on $$\lim_{N\to\infty}\text{Pr}(\frac{Y_N}{N}\leq \alpha)$$ for $0<\alpha<1$
On
Let $S_{n}=\sum_{j=1}^{n} X_{j}$.
The number of different walks of n steps where each step is +1 or −1 is $2^{n}$. For the simple random walk, each of these walks is equally likely. In order for $S_{n}$ to be equal to a number k it is necessary and sufficient that the number of +1 in the walk exceeds those of −1 by k. It follows +1 must appear \frac{n + k}{2} times among n steps of a walk, hence the number of walks which satisfy $S_{n} = k$ equals the number of ways of choosing (n + k)/2 elements from an n element set. Notice that n+k must be even since n=j+k+j where j is the number of -1 steps, so $n+k = 2j + 2k$ and we get the formula.
$$P(S_{n}=k) = \frac{n!}{(n-\frac{n+k}{2})!(\frac{n+k}{2})!2^{n}}$$
Using a symmetry we get $P(S_{n}=k)=P(S_{n}=-k)$.
So for n and k even:
$$P(|S_{n}|\leq k) = \sum^{k/2}_{j=0}2 P(S_{n} = 2j) - P(S_{n} = 0) = - \frac{n!}{2^{n}(\frac{n}{2}!)^{2}} + 2\sum^{k/2}_{j=0} \frac{n!}{(n-\frac{n+2j}{2})!(\frac{n+2j}{2})!2^{n}}$$
We can use a bound from https://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n
To get the bound $$P(|S_{n}|\leq k) \leq \frac{{n}\choose{k/2 +1}}{{n}\choose{n/2}} - \frac{{n}\choose{n/2}}{2^{n}}$$
Let $S_n=X_1+\dots+X_n$ be your position on the random walk.
The key claim is that $\mathbb P(Y_n\geq y)=\mathbb P(\lvert S_n\rvert\geq y)+\mathbb P(\lvert S_n\rvert\geq y+1)$.
First, note that if $s\geq y$, we have $\mathbb P(Y_n\geq y, \lvert S_n\rvert=s)=\mathbb P(\lvert S_n\rvert=s)$. Now if $\lvert S_n\rvert=s<y$, then by the reflection principle, $$\mathbb P(Y_n\geq y, \lvert S_n\rvert=s)=\mathbb P(Y_n\geq y, \lvert S_n\rvert=2y-s)=\mathbb P(\lvert S_n\rvert=2y-s).$$ So it follows that \begin{align*} \mathbb P(Y_n\geq y) &= \sum_{s}\mathbb P(Y_n\geq y, \lvert S_n\rvert=s) \\ &= \sum_{s\leq y-1}\mathbb P(\lvert S_n\rvert=2y-s)+\sum_{s\geq y}\mathbb P(\lvert S_n\rvert=s) \\ &=\mathbb P(\lvert S_n\rvert\geq y+1)+\mathbb P(\lvert S_n\rvert\geq y), \end{align*} as claimed. Now we can use the law of large numbers to finish: $$\lim_{N\to\infty}\mathbb P(Y_N/N<\alpha)=1-2\lim_{N\to\infty}\mathbb P(\lvert S_n\rvert/N\geq\alpha)=1.$$