Stationary probability in an M/M/$1$ queue with a lazy server 
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?
 A: 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 
$$\dfrac{1/\lambda}{n/\lambda + n/(\mu - \lambda)} = \dfrac{\mu - \lambda}{n\mu}$$
A: If I'm not mistaken, divide through $\lambda + \mu$ to get the probability, then use the fact that that $\frac{\lambda}{\lambda + \mu} + \frac{\mu}{\lambda +\mu} = 1$ and construct difference equations. Can you handle from here? 
