The Star Trek Problem in Williams's Book

$R^2_{n+1}/R^2_n$ has the same distribution as $|u+V|^2$ where $u$ is a fixed unit vector in ${\mathbb R}^3$ and $V$ is a random unit vector (uniform on the unit sphere). I would guess (although I haven't computed it) that $E [\log (|u+V|^2)] > 0$, and the Central Limit Theorem will tell you that almost surely $R_n > c^n$ for some $c > 1$. (EDIT: for sufficiently large $n$; Strong Law of Large Numbers is actually more relevant).

EDIT: Yes, $E[\log(|u+V|^2)] = \int_0^{\pi} \sin(\theta) \log(2 - 2 \cos(\theta))\ d\theta/2 = 2 \log(2) - 1 > 0$.

In the two-dimensional version, however, $E[\log(|u+V|^2)] = \int_0^\pi \log(2 - 2 \cos(\theta))\ d\theta/\pi = 0$, so the Law of Large Numbers doesn't tell you whether $R_n$ will go to $0$ or $\infty$. However, the Central Limit Theorem will say $P(R_n < 1) \to 1/2$ as $n \to \infty$. Thus it's certainly not the case in the two-dimensional version that $R_n \to 0$ with probability $1$, nor does $R_n \to \infty$ with probability $1$.

I think this way works basically, but there are some details to be completed.

Let $A_k$ be the event that the starship goes back to the Solar System just after the n th space jump. Then $\{A_k\}_{k=1}^\infty$ is a sequence of disjoint events. Obviously $A_k$ is adapt to $\mathcal{F}_n$, here $\mathcal{F}_n$ is the $\sigma$- field generated by the first $n$ jumps.

Let $p_{i+1}=P(A_{i+1}|\mathcal{F}_{i})$, then$$p_{i+1}= \frac{R_n-\sqrt{R^2_n-r^2}}{2R_n}$$

this follows from the area formula of a sphere cap.

(Here I am not sure because $ \frac{R_n-\sqrt{R^2_n-r^2}}{2R_n}$ is the conditional probability that given the ship is outside the solar system after n jumps, it will go back home after the next (n+1 th) jump, is it the same with $P(A_{i+1}|\mathcal{F}_{i})$?)

suppose it is, then we use the inequality $$1-\sqrt{1-x}\geq \frac{x}{2}\quad (0<x<1)$$ to derive $$p_{n+1}\geq \frac{r^2}{4R^2_n}$$ so if $\sum_{n}p_n$ converges, so dose $\sum_{n}\frac{1}{R^2_n}$

Levy's extension of the Borel-Cantelli lemma says (see page 124 in Williams's book)

on the set $S=\{\omega:\sum_{n}p_n=\infty\}$, almost surely holds $$\frac{\sum_{k=1}^n \mathbf{1}_{A_k}}{\sum_{k=1}^n p_k}\rightarrow 1$$

but the numerator can only have values 0 and 1,(note the $A_k$s are disjoint) so the set $S$ must have measure 0.