Mean position of random walk with only $+1$ jumps, with probability $\frac{c^x}{t-x+1}$ at time $t$ and site $x$ Consider the following random walk in one dimension, starting from $r(0)=0$.
$$
r(i+1) = r(i) + \xi,
$$
where $\xi(i, r(i))$ is an increment with distribution $P(\xi=1) = \frac{c^{r(i)}}{i-r(i)+1}$ and $\mathbb{P}(\xi=0) = 1 -\mathbb{P}(\xi=1)$, with $c<1$ constant. 
The question is: does the expectation $\mathbb{E}[r(i)]$ goes to infinity as $i$ goes to infinity or does it converge to a constant? For instance could it be of order $O( \log(i) )$?

The process $(r_c(t))$ based on $P(\xi(t,n)=1)=c^n$ jumps from $n$ to $n+1$ after a geometric time with mean $1/c^n$ hence $(r_c(t))$ hits $n$ after $\Theta(1/c^n)$ steps. Since $E(r_c(t))=\Theta(\log t)$ and $r_c(t)\geqslant r(t)$, $E(r(t))=O(\log t)$.
On the other hand, $r(t)\to+\infty$ almost surely. Otherwise, $r(t)$ would stay at some level $n$ forever after some time $t\geqslant n$, which happens with probability
$$
\prod_{s=t}^\infty\left(1-\frac{c^n}{s-n+1}\right)=0.
$$
