I have the following discrete-time Markov chain defined on the state space $S:=\{0,1,2,\ldots\}$: $$p(i,j) = \begin{cases} 1, \quad &\text{if $i=0$ and $j=1$}\\ \frac{1}{2}, \quad &\text{if $j=i-1$ and $i=1,2,\ldots$}\\ \frac{i-1}{2(i+1)}, \quad &\text{if $j=i$ and $i=1,2,\ldots$}\\ \frac{1}{i+1}, \quad &\text{if $j=i+1$ and $i=1,2,\ldots$}\\ 0, \quad &\text{otherwise} \end{cases}$$
In other words, $$P = \begin{bmatrix} 0&1\\ 1/2 & 0 &1/2\\ & 1/2& 1/(2*3)&1/3\\ &&1/2&2/(2*4)&1/4\\ &&&1/2&3/(2*5)&1/5\\ &&&\ddots&\ddots&\ddots \end{bmatrix}$$
This question uses the same DTMC used in this question. Unfortunately I'm stuck with another piece of this problem.
I am asked to find the expected number of steps it will take to reach state 0 given that I start in state 1. I tried using a first-step analysis to solve this problem, but it creates an infinite system of equations that I don't know how to solve. Specifically, let $\tau_0$ be the random variable representing the number of steps needed to state 0. Let $\mathbb{E}_i[\cdot]= \mathbb{E}[\ \cdot \ | \ X_0 = i]$ and let $\mathbb{P}_i(\cdot) = \mathbb{P}(\ \cdot \ | \ X_0 = i)$.
\begin{align*} \mathbb{E}_1[\tau_0] &= \mathbb{P}_1(X_1 = 0)\mathbb{E}_0[\tau_0] + \mathbb{P}_1(X_1 = 2)\mathbb{E}_2[\tau_0]\\ &= \frac{1}{2} (1) + \frac{1}{2} (\mathbb{E}_2[\tau_0]+1) \end{align*}
Then If we try to calculate $\mathbb{E}_2[\tau_0]$, we get a function of $\mathbb{E}_3[\tau_0]$, and this will produce an infinite system of equations that I don't know what to do with.
EDIT: In the linked question above, the stationary distribution is found.