 Solution

Every upward transition has positive probability. From any positive state the transition directly to zero has positive probability, since $1-\lambda /(i+1)>0$. Thus any state can reach zero and then reach any other state by successive upward steps, proving irreducibility.

If $\lambda <1$, the holding probability at zero is positive, so its period and hence the common period of the irreducible chain is one. At $\lambda =1$ that holding probability vanishes, but zero has positive-probability return paths $0\to 1\to 0$ of length two and $0\to 1\to 2\to 0$ of length three. Their lengths have greatest common divisor one. Thus $\boxed{\text{the chain is irreducible and aperiodic for all }0<\lambda \le 1}$.