I want to show that the escape speed of simple random walk on $\mathbb{Z}^d$ is zero. Let $S_n=X_1+X_2+\cdots+X_n$ be a simple random walk starting from origin where $X_i$ are i.i.d. equipped with uniform distribution on $\{1,2,\ldots, d\}$. We define escape speed as $\ell:=\lim_{n\to\infty}\frac{|S_n|}{n}$ where $S_n$ is the word norm or simply the distance from origin. I want to show that $\ell=0$ for every $d$.
Here are my thoughts so far: (1) When $d=1,2$, the random walk is recurrent which implies $\ell=0$; (2) I am confused when $d\geq 3$. The hint is to use Strong Law of Large Number. I have trouble setting up the random variable $|S_n|$. It would be a pleasure if I can write $|S_n|=\sum_{i=1}^n Y_n+Z$ where $Y_n$ are increments i.i.d. with zero expectation and $EZ$ bounded. However, $Y_n$ depends on the position of $S_{n-1}$. For example, if $S_n=0$ then $Y_{n+1}=1$ has probability one while if $S_n$ lies in the $x$-axis the $Y_{n+1}=1$ has probability $\frac{2d-1}{2d}$. I have trouble compute the expectation of $Y_n$.
Any help would be appreciated!
The thing to notice is that if $a_n \to 0$, then $|a_n| \to 0$ as well. Let $S_n = (S_n^{(1)}, \dots, S_n^{(d)})$; clearly in every single dimension $\lim_{n \to \infty} \frac{S_n^{(k)}}{n} = 0$ almost surely by the SLLN; but this means that $\lim_{n \to \infty} \frac{|S_n^{(k)}|}{n} = 0$ almost surely. Then $$ \lim_{n \to \infty} \frac{|S_n|}{n} \leq \lim_{n \to \infty} \sum_{k=1}^d \frac{|S_n^{(k)}|}{n} = 0 $$ happens with full probability which is what you want. Note that the walk being symmetric is key.