= Solution
{c}
First identify the recurrence type, rather than infer an occupation limit from a one-time <probability>. At time $k(a+b)$, return to zero requires exactly $kb$ upward steps. The <binomial distribution> gives
$$
\mathbb P_0(S_{k(a+b)}=0)=\binom{k(a+b)}{kb}\left(\frac b{a+b}\right)^{kb}\left(\frac a{a+b}\right)^{ka}
\sim\sqrt{\frac{a+b}{2\pi abk}},
$$
where the last step is <Stirling's approximation>. At all other positive times the return <probability> is zero. Hence $\sum_{n\geq0}\mathbb P_0(S_n=0)=\infty$. The <recurrence criterion by return probabilities> proves the <recurrent> property; by the <irreducible Markov chain> property established in (i), every state is <recurrent>. The equivalence of <positive recurrence> and existence of a <stationary distribution> for an <irreducible Markov chain> on a <countable> <state space>, together with (iii), makes this chain <null recurrent>.
For completeness, the zero occupation proportion follows directly by treating returns as a <renewal process>. Let $L_1,L_2,\ldots$ be successive strictly positive return-cycle lengths to zero. The <Strong Markov property> makes them <independent and identically distributed>, the <recurrent> property makes them finite <almost surely>, and the <null recurrent> property gives $\mathbb EL_1=\infty$. For each positive <integer> $K$, the <strong law of large numbers> applied to $\min(L_j,K)$ gives
$$
\liminf_{r\to\infty}\frac{L_1+\cdots+L_r}{r}\geq\mathbb E\min(L_1,K)\quad\text{almost surely}.
$$
Take a common event of <probability> one for these countably many $K$. By the <monotone convergence theorem>, the right side increases to infinity. Thus the number of completed return cycles by time $n$, divided by $n$, tends to zero. Counting the initial visit changes this ratio by at most $1/n$. Consequently,
$$
\boxed{\frac1n\sum_{r=0}^{n-1}\mathbf1_{\{S_r=0\}}\longrightarrow0\quad\text{almost surely}.}
$$
The walk visits zero infinitely often, but the asymptotic fraction of time spent there is zero.
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