Let $B_t$ be a standard Brownian motion. I have to show that $$\int_0^t \frac{B_u}{u}du$$ converges almost surely with $t\to \infty$. Honestly, I do not know how to proceed, since I am new to the field of stochastic processes, what I tried to do is to say that, for every $s$, we have $$ \int_0^t \frac{B_u}{u}du = \int_0^s \frac{B_u}{u}du + \int_s^t \frac{B_u}{u}du $$ but then, by definition of Brownian motion, $$\int_s^t \frac{B_u}{u}du = \int_s^t \frac{B_u-B_s}{u}+\frac{B_s}{u} du=\int_s^t \frac{B_u-B_s}{u}+ln(s)B_s$$ where as first term we have something similar to the initial integral, even if probably I have just made things more complex. Could you give me some hints?
EDIT:
There was a misunderstanding here, the statement proposed by the book where I've found the problem was instead referring to the convergence of the integral in the 0 extremal, which is much simpler. Thus, it is difficult to know wether this statement is true or not.