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For any vertex $v$ which is not in the all-left ray, there is some generation $n$ so that all descendants of $v$ in the $n$th generation are in the right half. (If the fraction of vertices to the left of $v$ in its level is $\alpha$, this happens by $\lceil \log_{3/2} 1/(2\alpha) \rceil$ generations below $v$, since the fraction of vertices to the left expands by a factor of $3/2$ until the generation when it reaches $1/2$.) This means the number of descendants of $v$ in each generation is eventually constant, so the resistance in the downward direction is infinite. Therefore, and with probability $1$ a random walk starting at $v$ will reach the parent of $v$.

So, a random walk on the $3-1$ tree almost surely retracts to a random walk on the all-left ray. Since this random walk is recurrent, the random walk on the $3-1$ tree is recurrent.

For any vertex $v$ which is not in the all-left ray, there is some generation $n$ so that all descendants of $v$ in the $n$th generation are in the right half. (If the fraction of vertices to the left of $v$ in its level is $\alpha$, this happens by $\log_{3/2} 1/(2\alpha)$ \lceil \log_{3/2} 1/(2\alpha) \rceil$generations below$v$, since the fraction of vertices to the left expands by$3/2$until the generation when it reaches$1/2$). 1/2$.) This means the number of descendants of $v$ in each generation is eventually constant, so the resistance in the downward direction is infinite, and with probability $1$ a random walk starting at $v$ will reach the parent of $v$.
So, a random walk on the $3-1$ tree retracts to a random walk on the all-left ray. Since this random walk is recurrent, the random walk on the $3-1$ tree is recurrent.
For any vertex $v$ which is not in the all-left ray, there is some generation $n$ so that all descendants of $v$ in the $n$th generation are in the right half. (If the fraction of vertices to the left of $v$ in its level is $\alpha$, this happens by $\log_{3/2} 1/(2\alpha)$ generations below $v$, since the fraction of vertices to the left expands by $3/2$ until it reaches $1/2$). This means the number of descendants of $v$ in each generation is eventually constant, so the resistance in the downward direction is infinite, and with probability $1$ a random walk starting at $v$ will reach the parent of $v$.
So, a random walk on the $3-1$ tree retracts to a random walk on the all-left ray. Since this random walk is recurrent, the random walk on the $3-1$ tree is recurrent.