Infinite summation converges or not

Question

Let $\{X_n\}$ be an i.i.d. random sequence, and satisfies $$\mathbb{P}(X_n = 1) = \mathbb{P}(X_n = -1) = \frac{1}{2}.$$ Then I am curious about the probability of the following event: $$\left\{ \sum_{n=1}^\infty 2^{\sum\limits_{k=0}^{n-1}X_k}<\infty \right\}.$$ As we see, this is a well defined event. However, I cannot come across any theorems in probability theory to give an estimation or bound to this one (even "$\mathbb{P}$>0" or "$\mathbb{P}$<1").

Noted that, here we can enjoy the property of strong law of large numbers, but we cannot use the root test for the positive series, since $$\lim_{n\to\infty}(2^{\sum\limits_{k=0}^{n-1}X_k})^{1/n} = 1.$$

Any discussion is welcome.

Answer 1

The symmetric random walk $S_n:=\sum_{k=0}^{n-1}X_k$ is known to “oscillate” almost surely. In particular, almost surely, $S_n\ge1$ for infinitely many $n$: there exists an event $\Omega$ such that $\Bbb P(\Omega)=1$ and for all $\omega\in\Omega$, $S_{n_k(\omega)}\ge1$ along some increasing sequence of integers $(n_k(\omega))_{k\ge1}$. Thus $$\sum_{n=1}^\infty 2^{S_n(\omega)}\ge\sum_{k=1}^\infty2^{S_{n_k(\omega)}(\omega)}\ge\sum_{k=1}^\infty2^1=\infty$$ will hold almost surely.