# Infinite summation converges or not

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.

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.