= Solution
Every step changes the population by one, so $X_n$ has the parity of $n$ when $X_0=0$. Hence $\mathbb P(X_{2m+1}=0)=0$. A return in two steps has positive <probability> from every state, and all return times are even, so the <period of a state in a Markov chain> is exactly two.
Each transition exchanges the even and odd classes. Since $\pi$ is an <invariant distribution>, their masses are equal: $\pi(\text{even})=\pi(\text{odd})=1/2$. The chain sampled at times $2m$, restricted to the even states, is irreducible and positive recurrent, and it is aperiodic because its diagonal entries are positive. Its <stationary distribution> is $2\pi_k$ on even $k$. The <countable-state Markov chain convergence theorem> applied to this sampled chain yields
$$
\boxed{\mathbb P(X_{2m}=0)\longrightarrow2\pi_0=(1-r)^{a+1},\qquad \mathbb P(X_{2m+1}=0)=0.}
$$
Since $\pi_0>0$, the full sequence has no limit. Its time average nevertheless tends to $\pi_0$ by the <countable-state Markov chain ergodic theorem>.
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