= Solution
Every finite upward path has positive probability, since each upward transition probability is positive. Every state $i\geq1$ has a positive transition probability to zero because $\lambda/(i+1)<1$. From zero one can therefore reach every state, and from any state return to zero and then reach any other. Hence the chain is <irreducible>.
If $\lambda<1$, state zero has a positive self-loop, so its period is one. If $\lambda=1$, that self-loop disappears, but the paths $0\to1\to0$ and $0\to1\to2\to0$ have positive probabilities and lengths two and three. Their greatest common divisor is one. Thus zero has period one in this boundary case too; irreducibility makes the whole chain \b[<aperiodic>].
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