Recurrence/Transience of random walk with +2/-1 steps Consider the Markov chain with state space $S=(0,1,2,...)$ and transition probabilities: $p(x,x+2)=p$ , $p(x,x-1)=1-p$, $\forall$ $x>0$. $p(0,2)=p$ , $p(0,0)=1-p$.
For which values of $p$ is this a transient chain?
We know that the Markov chain is transient if $\sum_{n=0}^{\infty}p_{n}(x,x)< \infty$ 
i.e the expected number of returns to a state is finite. I tried thinking about the expected number of returns to state zero and I know that the number of jumps to the right has to be half the number of moves to the left. But I cannot figure out the permutations for this.

Consider the Markov chain $(X_n)$ starting from $1$ and $T$ the hitting time of $0$, possibly infinite, then $(X_n)_{n\leqslant T}$ is distributed like the random walk $(Y_n)_{n\leqslant S}$ on $\mathbb Z$ with steps $+2$ and $-1$ of respective probabilities $p$ and $1-p$, starting from $Y_0=1$, with $S$ the first hitting time of $0$ by $(Y_n)$, possibly infinite.
Thus $P(T\lt\infty)=P(S\lt\infty)$. Since $P(S\lt\infty)=1$ if and only if the drift $E(Y_1-Y_0)$ of $(Y_n)$ is nonpositive, the Markov chain $(X_n)$ is recurrent if and only if $p\leqslant\frac13$.
