Gambler's ruin expected mean time of absorption

Question

I'm not sure whether this is right or not. Consider this discrete-time Markov Chain $(X_{n})_{n \ge 0}$ with infinite state space $E = \{0,1,2,...\}$ with transition probability matrix $P$. $$ P = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & \cdots \\ 0.5 & 0 & 0.5 & 0 & 0 & \cdots \\ 0 & 0.5 & 0 & 0.5 & 0 & \cdots \\ 0 & 0 & 0.5 & 0 & 0.5 & \cdots \\ \vdots & & & & & \ddots \end{pmatrix} $$ Let $T = \{ min ( n \ge 1 | X_{n} \in \{0\} )\}$ and $w_{i} = \mathbb{E}[T | X_{0} = i]$ for $i = 1,2,...$
Using: $$ w_{i}= 1 + \sum_{j \in \{1,2,...\}} P_{ij}w_{j} $$ I'm getting: $$ w_{1} = 1 + \frac{1}{2}w_{2} $$ $$ w_{i} = 1 + \frac{1}{2}w_{i-1} + \frac{1}{2}w_{i+1} \quad \quad i =2,3,.. $$ The solution of the associated homogeneous equation is $w_{i} = A + Bi$, so the solution of the non homogeneous equation will be $ w_{i} = A + Bi + u_{i} $ , where $u_{i}$ is the particular/special solution. One can find $u_{i} = -i^2 $.
So $w_{i} = A + Bi -i^2$. Setting $w_{0} = 0$ one gets $A = 0$. But once arrived here i do not know what else to do, my intuition tells me that: $$ \lim_{i \to +\infty} w_{i} = + \infty $$ Does this mean that $w_{i} = + \infty $ for $i = 1,2,...$ ?