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All specified nearest-neighbour probabilities are positive, making the birth-death chain irreducible. The row sums give $p_0+q_0=1$ and $p_i+q_i=1$ for $i\ge 1$. Write $T_j$ for the hitting time of $j$, and on $\{0,\dots ,N\}$ let $h_i={\mathrm{Pr}}_i(T_N<T_0)$. The first-step equations giveDefine ${\rho }_0=1$ and ${\rho }_k={\prod }_{r=1}^k{q}_r/p_r$. The successive increments are proportional to ${\rho }_k$, and normalization at $N$ yieldsThe chain cannot remain forever in a fixed finite set avoiding zero: from every state in that set there is a uniformly positive chance to leave or hit zero within finitely many steps, so iterating the Markov property makes indefinite confinement have probability zero. Thus avoiding zero is the limit event of hitting every upper boundary before zero. Sending $N\to \infty$ shows that the probability of never hitting zero from one vanishes exactly when ${\sum }_k{\rho }_k=\infty$. At zero, return is immediate with probability $q_0$ and otherwise begins an excursion from one; irreducibility then givesFor positive recurrence define $w_0=1$ and $w_n={\prod }_{r=1}^n{p}_{r-1}/q_r$. First assume recurrence and let $\tau =T_0^{+}$ be the first return from zero. For each fixed $i$, the expected number $m_i$ of visits to $i$ before $\tau$ is finite: each visit to $i\ge 1$ has a positive chance of following the successive downward steps to zero before another visit, so its visit count has a geometric tail. Also $m_0=1$.

Each finite excursion starts and ends below every edge $(i,i+1)$, so its upward and downward crossing counts of that edge are equal. Their expectations, by the Markov property, are $m_i{p}_i$ and $m_{i+1}{q}_{i+1}$. Hence $m_{i+1}=m_i{p}_i/q_{i+1}$ and $m_i=w_i$. Tonelli's theorem applied to the nonnegative occupation counts gives the excursion weights of a reflected birth-death chain formulaConsequently positive recurrence implies the requested finite sum. Conversely, if ${\sum }_i{w}_i<\infty$, then $w_i\to 0$. But ${\rho }_i{w}_i=p_0/p_i\ge p_0$, so ${\rho }_i\ge p_0/w_i\to \infty$ and the already proved recurrence criterion holds. The occupation calculation is therefore valid and gives a finite mean return time. ThusAdding the finite $w_0=1$ term does not change convergence.