Question on part 3 of the Star Trek problem in Williams, Probability with Martingales 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.
 A: 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.
