= Excursion weights of a reflected birth-death chain
{title2=$w_n=\prod_{r=1}^n p_{r-1}/q_r$}
Consider an <irreducible Markov chain>[irreducible] <Birth-death chain> on $\{0,1,2,\ldots\}$. At zero let $P(0,1)=p_0>0$ and $P(0,0)=1-p_0$. At $i\geq1$, let $P(i,i+1)=p_i>0$ and $P(i,i-1)=q_i>0$, with $p_i+q_i=1$. Define
$$
w_0=1,\qquad w_n=\prod_{r=1}^n\frac{p_{r-1}}{q_r},\qquad
\tau_0^+=\inf\{m\geq1:X_m=0\}.
$$
In the <recurrent Markov chain>[recurrent] case, $w_i$ is the expected number of visits to $i$ during a cycle started at zero, counting times $0\leq m<\tau_0^+$. The initial visit gives $w_0=1$. Each upward crossing of an <edge> before the return is paired with a downward crossing, giving $w_ip_i=w_{i+1}q_{i+1}$ and the product formula. Tonelli's theorem then gives $\mathbb E_0\tau_0^+=\sum_iw_i$. Consequently the <recurrent Markov chain>[recurrent] chain is <positive recurrent Markov chain>[positive recurrent] exactly when this sum is finite, and its <Stationary distribution> is $w_i/\sum_jw_j$.
The finiteness criterion also rules out transience without assuming recurrence in advance. Put $\rho_n=\prod_{r=1}^nq_r/p_r$ and $\rho_0=1$. Then $w_n\rho_n=p_0/p_n\geq p_0$. If $\sum_nw_n<\infty$, then $w_n\to0$, so $\rho_n\to\infty$ and $\sum_n\rho_n=\infty$. The birth-death scale-function hitting formula therefore implies recurrence. This separates stationary occupation weights $w_n$ from the inverse-direction scale weights $\rho_n$, which serve different purposes.
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