= Solution
When $\mu=\lambda$, an excursion after the jump from zero to one visits $n$ before its reset with probability
$$
\prod_{j=1}^{n-1}\frac{j+1}{j+2}=\frac2{n+1}.
$$
This tends to zero, so a catastrophe eventually occurs with probability one. Both the original chain and its <jump chain> are recurrent.
The expected number of jumps in a return cycle is
$$
1+\sum_{n\ge1}\frac2{n+1}=\infty.
$$
Thus the <jump chain> is \b[null recurrent]. But its holding time at $n\ge1$ has mean $1/[\lambda(n+2)]$, and the initial holding time at zero has mean $1/\lambda$. The expected elapsed cycle time is
$$
\frac1\lambda+\sum_{n\ge1}\frac2{(n+1)\lambda(n+2)}
=\frac2\lambda<\infty.
$$
The <continuous-time Markov chain> is therefore \b[positive recurrent]. Its equilibrium distribution, computed as mean time in a state per mean cycle time, is
$$
\boxed{\pi_0=\frac12,\qquad
\pi_n=\frac1{(n+1)(n+2)}\quad(n\ge1).}
$$
The entries sum to one by telescoping. The contrast is explained by progressively shorter holding times at the large states, not by a change in return probabilities.
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