There are many threads and answers on the Proof of the Strong Law of Large numbers. But my question is a specific example in these notes by Craig A. Tracy. The relevant parts are (page 3):
let $X_1, X_2, X_3, \dots$ denote an infinite sequence of independent random variables with common distribution. Set $S_n = X_1 + \dots + X_n$. […] (Take, for instance, in coining tossing the elementary event $\omega = HHHH\ldots$ for which $S_n(\omega) = 1$ for every $n$ and hence $\lim_{n\to\infty} S_n(\omega)/n = 1$.)
My question is: Is $S_n(\omega) = 1$ wrong?
Analysis: I guess $\omega$ denotes the event of constant, consecutive heads (as opposed to tails).
$$ S_n(\omega) = X_1(\text{HH}\ldots) + X_2(\text{HH}\ldots) + \ldots + X_n(\text{HH}\ldots) $$
The probability of a single heads result is $\frac12$. The probability of $k$ heads results is $\frac1{2^k}$. $S_n(\omega)$ should therefore be $n \cdot \frac1{2^k}$. This is different from his claim $S_n(\omega) = 1$.
If we assume $S_n(\omega) = 1$, I think the limes is also wrong.
$$ \lim_{n \to \infty} \frac{S_n(\omega)}{n} = \lim_{n\to\infty} \frac{1}{n} = 0 \neq 1 $$
But my claim $S_n(\omega) = n \cdot \frac{1}{2^k}$, is not correct either:
$$ \lim_{n \to \infty} \frac{S_n(\omega)}{n} = \lim_{n\to\infty} \frac{n \cdot \frac{1}{2^k}}{n} = \frac{1}{2^k} \neq 1 $$
If we assume $k = n$, we get $S_n(\omega) = 0$ as well. $k$ must be $0$ to hold true, which would be pointless. So where am I wrong here?