Markov Chains to find total reward I have the following state space corresponding to the subscription details of customers with number of months: $S=\{C,0,1,2,3,4,5,6,7,8,...\}$. This corresponds to the number of months of subscription and $C$ means cancelled. 
Can I use stationary probabilities to model this? I think it has to be a positively recurrent process, to find stationary probabilities. Since in the long run, all customers will leave and end up at $C$. 
What I tried:
Let $p_{i,i+1}$ be the probability of going from state $i$ to $i+1$.
Let $q_{i,C} = 1 - p_{i,i+1}$
$p_{C,C}=1$
I'm trying to find out the value of each customer. I formulated the problem as a total reward problem when starting from state $0$, where $r=10$ is the reward at every month. Then expected total reward for state $0$ is:
$$\phi_i=10+q_{i,C}\phi_{C}+p_{i,i+1}\phi_{i+1} \text{ for } i\geq 0$$ and
$$\phi_C=0$$
However, ($p_{i,i+1}=0.9$ for $i\geq 0)$ I can't seem to solve this system of equation to get $\phi_0$.
 A: Let's cut the chain at level $i=N$ to get a finite Markov chain and set transition from the state $N$ to $C$ with probability $1$. Now this is a finite Markov chain with an absorbing state $C$. With results that can be found for example here (see page 418) we have that the expected number of times the chain is at state $j$ starting from the state $i$ (before being absorbed to the state $C$) is given by the fundamental matrix entry $n_{ij}$. The fundamental matrix is $N = (I-Q)^{-1}$ where $Q$ is the non-absorbing part of the transition matrix (see the link).
If we assume the transition probablility is constant $p_{i,i+1}=p$, we have
$$I-Q=\begin{bmatrix}
    1 & -p &  &   &  \\
     & 1 & -p &   &  \\
     &  & \ddots & \ddots &  \\
     &  &  &   & 1
\end{bmatrix}$$
It can be seen that the inverse of this is
$$N = \begin{bmatrix}
    1 & p & p^2 & \dots  & p^N \\
    0 & 1 & p & \dots  & p^{N-1} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & 0 & \dots  & 1
\end{bmatrix}$$
So when starting from the state $0$ the expected number of times the chain visits any state before $C$ is
$$\sum_{k=0}^N p^k.$$
If for each state we get the reward $r$ then the expected reward is $r\sum_{k=0}^N p^k$. It's at least intuitively clear that cutting the chain at larger and larger $N$ leads to an answer that is closer and closer to the infinite case. As we let $N\to\infty$ this geometric sum approaches $\frac{r}{1-p}$.
