= Solution
Every reachable state $i$ reaches zero deterministically in $i$ steps. For $j\in E$, choose a supported lifetime $k>j$; zero can jump to $k-1$ and then count down to $j$. Thus the chain is an <irreducible Markov chain> on $E$. Its positive-probability first return lengths are exactly the support of $Y_1$; all other return lengths are sums of such lengths. Their <greatest common divisor> is therefore one by assumption, so the chain is an <aperiodic Markov chain>.
The <equilibrium residual-life distribution> is
$$
\pi_j=\frac{\mathbb P(Y_1>j)}\mu,\qquad j\in E.
$$
It is a probability distribution because the tail-sum identity gives $\sum_{j\ge0}\mathbb P(Y_1>j)=\mathbb EY_1=\mu$. With zero masses outside $E$,
$$
(\pi P)_j=\pi_{j+1}+\pi_0\mathbb P(Y_1=j+1)
=\frac{\mathbb P(Y_1>j+1)+\mathbb P(Y_1=j+1)}\mu=\pi_j,
$$
so it is a <stationary distribution>. This also displays why finite mean is needed for normalization.
Use the <countable-state Markov chain convergence theorem>: an irreducible, aperiodic, positive recurrent discrete-time chain on a countable state space has a unique stationary distribution and $P^n(i,j)\to\pi_j$ for every pair of states. All hypotheses were verified, including positive recurrence in part (b). Since $X_0=0$ and $\mathbb P(Y_1>0)=1$,
$$
\boxed{\lim_{n\to\infty}\mathbb P(X_n=0)=\pi_0=\frac1\mu.}
$$
The finite-state convergence theorem alone would not suffice when the lifetimes are unbounded. The gcd hypothesis is also necessary for this ordinary limit: deterministic lifetime 2, for example, produces alternating renewal probabilities rather than convergence.
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