= Solution
Write a first return time as $2J$, where $\mathbb P(J=j)=2^{-j}$ for $j\geq1$. It has mean $2\cdot2=4$ and period two, so the stationary <probability> at one is $1/4$. The periodic convergence theorem would give limits $1/2$ along even times and zero along odd times.
Here one can obtain them directly. Put $u_n=\mathbb P_1(X_{2n}=1)$, with $u_0=1$. Splitting according to the first return gives the <renewal equation>
$$
u_n=\sum_{j=1}^n2^{-j}u_{n-j}.
$$
Induction gives $u_n=1/2$ for every $n\geq1$: the right side is $2^{-n}+\tfrac12\sum_{j=1}^{n-1}2^{-j}=1/2$. Odd-time returns are impossible because every return interval is even. Therefore the exact set of limit points is
$$
\boxed{\{0,\tfrac12\}.}
$$
The <sequence> does not converge to the stationary mass $1/4$, precisely because it is periodic.
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