This is a part of an exercise in Durrett's probability book.
Consider the Markov chain on $\{1,2,\cdots,N\}$ with $p_{ij}=1/(i-1)$ when $j and $p_{ij}=0$ otherwise. Suppose that we start at point $k$. We let $I_j=1$ if $X_n$ visits $j$. Then $I_1,I_2,\cdots,I_{k-1}$ are independent.
I don't find it obvious that $I_1,\cdots,I_{k-1}$ are independent. It is possible to prove the independence if we calculate all $P(\cap_{j\in J\subset\{1,\cdots,k-1\}}I_j)$, but this work is long and tedious. Since the independence was written as an obvious thing in this exercise, I assume that there is an easier way.