 Solution

Write a first return time as $2J$, where $\mathbb{P}(J=j)=2^{-j}$ for $j\ge 1$. It has mean $2\cdot 2=4$ 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 equationInduction gives $u_n=1/2$ for every $n\ge 1$: the right side is $2^{-n}+\frac{1}{2}{\sum }_{j=1}^{n-1}{2}^{-j}=1/2$. Odd-time returns are impossible because every return interval is even. Therefore the exact set of limit points isThe sequence does not converge to the stationary mass $1/4$, precisely because it is periodic.