Customers arrive to a single server queue according to a Poisson process with rate $\lambda$. Each customer requires Exponential($\mu$) service time. In the beginning when there are $0$ customer, the server idles (i.e. ignores the customers) until the number of customers reaches $n$. As soon as there are at least $n$ customers, the server begins working and serves every customer until the system is completely cleared (i.e. until there are zero customers). Then the server idles again until the number of customers reaches $n$ again.
Find the stationary probability that the system has zero customer.
Define state space $S=\mathbb{N}_0 \cup \{1_u,2_u,\ldots, (n-1)_u\}$. The states with subscript $u$ are the idle states where the server does not serve. I model this system as a continuous-time Markov chain illustrated by the following state diagram:
Let $p_s$ denote the stationary probability of the process being in state $s\in S$. Since $p_{1_u}\lambda=p_1\lambda$ and $p_{i_u}\lambda=p_{(i-1)_u}\lambda$ for $i=2,\ldots,n-1$, we have $p_{i_u}=p_0$ for $i=1,\ldots,n-1$. Then I can write out the balance equations in terms of only the $p_i, i\in\mathbb{N}_0$:$$ \begin{aligned} p_0\lambda &= p_1\mu \\ p_1 (\lambda+\mu) &= p_2 \mu \\ p_2 (\lambda+\mu) &= p_1 \lambda + p_3 \mu \\ &\vdots \\ p_i (\lambda+\mu) &= p_{i-1} \lambda + p_{i+1} \mu \\ &\vdots \\ p_{n-1}(\lambda +\mu) &= p_{n-2} \lambda + p_n \mu \\ p_n(\lambda+\mu) &= p_0 \lambda + p_{n-1}\lambda + p_{n+1} \mu \\ p_{n+1}(\lambda+\mu) &= p_{n}\lambda + p_{n+2} \mu \\ &\vdots \\ p_{j}(\lambda+\mu) &= p_{j-1}\lambda + p_{j+1} \mu \\ &\vdots \end{aligned} $$
I have been stuck from this point on. I did try writing out $p_1$ in terms of $p_0$, and $p_2$ in terms of $p_1$ (and hence $p_0$), and so on, trying to find an inductive pattern. I tried up to $p_5$, but still cannot see any clear pattern. Could anyone give some help?
Consider a "renewal time" as a time when the number of customers goes from $1$ to $0$. After a renewal time, we wait Exponential($\lambda$) time in each of the states $0, 1_u, \ldots, (n-1)_u$: a total expected time $n/ \lambda$ of which an expected time $1/\lambda$ is spent in state $0$. Then the server becomes active, and we are in a "normal" M/M/1 queue until the customers are cleared (presumably $\lambda < \mu$ so this will actually happen). The expected time for that is $n/(\mu - \lambda)$. So the total expected time between renewals is $n/\lambda + n/(\mu - \lambda)$, of which an expected time $1/\lambda$ is spent in state $0$. Thus the equilibrium probability of state $0$ is $$\dfrac{1/\lambda}{n/\lambda + n/(\mu - \lambda)} = \dfrac{\mu - \lambda}{n\mu}$$