Consider this M.SE question, which is E12.3 in Williams. The answer of Robert Israel (and Xoff) seems to give an exponential bound on $R_n$ almost surely. Wouldn't this imply the convergence of $$\sum R_n^{-1},$$ which is significantly stronger than what the problem asks to prove, which is the convergence of $$\sum R_n^{-2}?$$
I would like confirmation that this stronger result is actually true and that I am not missing anything.
The position of the ship after $n$ space-hops is${}^*$ $X_{n} = X_{n-1} + R_{n-1} U_n$, where the vector $V_n$ is independent of $X_1,\dots,X_{n-1}$ and uniformly distributed on the unit sphere $S_1$. So $R_n = R_{n-1}|e_{n-1}+V_n|$, where $e_{n-1}$ is a unit vector in the direction of $X_{n-1}$. Then, clearly, $R_n= R_{n-2}|e+U_{n-1}||e+U_n| = R\prod_{k=1}^n |e+U_k|,$ where $e$ is a fixed unit vector, and $U_1,U_2,\dots,U_k$ are independent and uniformly distributed on $S_1$. In order to study the asymptotic behavior of $R_n$, it is easier to consider its logarithm: $$ \log R_n = \log R + \sum_{k=1}^n \log |e+U_k|. $$ The latter sum consists of independent identically distributed random variables. Therefore, in view of the strong law of large numbers, $$ \frac1n \log R_n \to E[\log |e+U_1|], n\to\infty,\tag{1} $$ almost surely. Now $$ E[\log |e+U_1|] = \frac{1}{4\pi}\int_{S_1} \log|e+v| d\sigma(v); $$ taking $e = (0,0,-1)$ and parametrizing $S_1$ by $v = (\cos \varphi \sin \theta, \sin\varphi \sin \theta, \cos\theta)$, we get $|e+v| = \sqrt{\sin^2\theta + (\cos\theta-1)^2} = \sqrt{2-2\cos\theta}$, whence $$ E[\log |e+U_1|] = \frac{1}{4\pi}\int_0^{2\pi}\int_0^\pi \frac12 \sin\theta\log(2-2\cos\theta)d\theta\,d\varphi \\ = \frac14 \int_{-1}^1 \log(2+2x)dx = \log 2 - \frac12>0. $$ Therefore, in view of (1), for any $c\in(0,\log2-1/2)$, $\log R_n> cn$ eventually with probability 1, equivalently, $R_n> e^{cn}$. Therefore, the series $\sum_{n=1}^\infty \frac{1}{R_n}$ converges almost surely.