I'm attempting Exercise 5.33 of Le Gall's Brownian motion, Martingales and Stochastic Calculus.

Let $B_t$ be a 3-dimensional Brownian motion starting from $x$.

Part 6 asks me to show that $$|B_t| = |x| + \beta_t +\int_0^t\dfrac{ds}{|B_s|} \quad (*)$$

where $$\beta_t = \sum_{i=1}^3 \int_0^t \dfrac{B^i_s}{|B^i_s|} dB^i_s$$.

I have shown this by Ito's formula, and I have shown that $\beta_t$ is a 1-dimensional Brownian motion (by calculating the quadratic variation and using Levy's characterisation).

Now part 7 says

Show that $|B_t| \rightarrow \infty$ as $t \rightarrow \infty$ a.s. (Hint observe that $|B_t|^{-1}$ is a non-negative supermartingale.)

So I'm not sure show to show the transience. I feel like it could follow from (*), since if $|B_t|$ does not tend to infinity then the integral on the right hand side must tend to infinity, but then $|B_t|$ must tend to infinity to balance this (not sure how to make this rigorous).

My other idea is to say since $|B_t|^{-1}$ is an $L^1$ bounded supermartingale it converges a.s., and to show $|B_\infty|^{-1} = 0$, but again, I'm not sure exactly how to do this.

Thanks in advance.