= Solution
For each adjacent pair, the <detailed balance equations> follows directly:
$$
\pi_nP(n,n+1)=\tfrac12\pi_n\min(1,\pi_{n+1}/\pi_n)=\tfrac12\min(\pi_n,\pi_{n+1})=\pi_{n+1}P(n+1,n).
$$
For nonadjacent distinct states both fluxes vanish, and diagonal fluxes agree trivially. Since $\sum_n\pi_n=e^{-\lambda}\sum_n\lambda^n/n!=1$, this is a genuine <stationary distribution>, rather than merely an invariant measure of infinite mass.
Every upward and downward <transition probability> between neighbouring states is positive for $\lambda>0$. Any two states can therefore be joined by a finite path of positive <probability>: the <Markov chain> is <irreducible>. Moreover,
$$
P(0,0)=1-\tfrac12\min(1,\lambda)\geq\tfrac12>0.
$$
State zero has period one; <Markov chain irreducibility> makes every state <aperiodic>. Finally, the standard countable-state recurrence theorem states that an <irreducible Markov chain> admitting a <stationary distribution> of total mass one is <positive recurrent>, with that <stationary distribution> unique. The <countable-state Markov chain convergence theorem> then states that <Markov chain irreducibility>, <positive recurrence> and <aperiodicity> imply $P^k(i,j)\to\pi_j$ for all states $i,j$. The hypotheses just verified hold on the infinite <state space> $\mathbb N_0$, not merely on a numerical truncation.<Markov Chains by James R. Norris>[Sections 1.7–1.8, invariant distributions and convergence to equilibrium]{ref}
Consequently, \b[the unique <stationary distribution> and the <limiting distribution> are both $\operatorname{Poisson}(\lambda)$], from every initial state. The <countable-state Markov chain ergodic theorem> also gives $N^{-1}\sum_{k=1}^Nf(X_k)\to\sum_n\pi_nf(n)$ <almost surely> whenever $\sum_n\pi_n|f(n)|<\infty$.<Markov Chains by James R. Norris>[Section 1.10, ergodic theorem]{ref}
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