Behavior of explosive random process Inspired somewhat by this problem, I've been investigating the behavior under iteration of the following discrete random process:

Given $n\in\mathbb{N}$, choose an integer from $\{0,1,\ldots,n\}$ uniformly at random, and multiply $n$ by it.

The process a.s. either reaches its fixed point, $0$, or diverges to infinity.  (The only other possibility, that $n$ remains forever at some non-zero value, clearly has zero probability.)  When it diverges to infinity, it does so quickly: its average value grows super-exponentially with the number of steps.  I'm interested in the probability of the process reaching $0$, and in the cases where it does reach $0$, the largest value it attained.
Let $p(n)$ be the probability of reaching $0$ starting from $n$.  Then we have
$$
p(n) = \frac{1}{n+1}\left(1 + \sum_{k=1}^n p(kn)\right),
$$
or
$$
p(n) = \frac{1}{n} + \frac{1}{n}\sum_{k=2}^n p(kn).
$$
We have $p(n) \ge 1/n$ immediately; feeding this back in gives an improved lower bound of
$$
p(n) \ge \frac{1}{n}\left(1 + \sum_{k=2}^{n}\frac{1}{kn}\right) = \frac{1}{n}+\frac{1}{n^2}\left(H_{n}-1\right),
$$
where $H_n$ is the $n$-th harmonic number.  The lower bound can be improved by repeating this process; can a useful asymptotic expansion for $p(n)$, or even a closed form, be derived?
Similarly, let $q(n)$ be the expected value immediately preceding $0$ among terminating sequences that start from $n$.  It is clear from the definition that $q(n)\ge n$, and numerical simulation suggests that $q(n) \sim K n$ for some constant $K$.  Can $q(n)$ be given an asymptotic expansion?  What is the value of the constant K?
 A: I wrote the following comment (now corrected for having been $1-p(n)$):     
I would guess for large $n$, you almost never fail unless it is on the first try, so would guess that $p(n)\sim \frac 1n$ and therefore $K=1$. This is because for large n you almost surely multiply by a large number. But "almost" here is not (quite) "with probability 1"
If we take it too seriously and try to iterate, we would say $p(kn)=\frac 1{kn}$.  Putting that into your first equation we have $p(n)=\frac 1{n+1}\left(1+\sum_{k=1}^n \frac 1{kn}\right)=\frac 1{n+1}+\frac{H_n}{n(n+1)}\approx \frac 1{n+1}(1+\frac {\ln n}{n})$ which looks to be converging rapidly for large $n$.  This is still wild handwaving, but it might help.
In this picture, the probability that we never get greater than $n$ is simply the chance that we hit some number of $1$'s (maybe none) and then a zero.  This is $\sum_{i=1}^\infty \frac 1{n+1}=\frac 1n$.  The chance that we get greater than $n$ and then hit zero is all the rest, which is $\frac 1n-\frac 1{n+1}(1+\frac {\ln n}{n})=\frac {1+\ln n}{n(n+1)}$.  The chance that we hit $2n$ and no more is the chance that we hit some number of 1's, a 2, some more 1's, and a zero.  This is $(\sum_{i=0}^\infty\frac 1{n+1})\frac 1{n+1}(\sum_{i=0}^\infty\frac 1{2n+1})\frac 1{2n+1}=\frac 1{n(n+1)2n(2n+1)}$ which is already $n^{-4}$, so I think $\lim_{n \to \infty} \frac {q(n)}n=1$
