I have the following $M/M/1$ queue where:
$\cdot$ customers arrive at a rate $\lambda$ and they require an exponential service time with rate $\mu$.
$\cdot$ customers leave the queue at a rate $\delta$, independently of their position in the queue.
I am asked to prove that for any $\delta>0$ the queue has an invariant distribution.
My approach: I am trying to find the generator(Q-matrix) in order to apply detail balance equation.
The process of leaving the queue and the service process are independent so $g_{i,i-1}=\mu+\delta$. And as customers arrive as a Poisson process with rate $\lambda$ we have that $g_{i,i+1}=\lambda$.
Applying detail balance we have that:
$\pi_{i}g_{i,i-1}=\pi_{i-1}g_{i-1,i}$
$\pi_{i}=\frac{\lambda}{\mu+\delta}\pi_{i-1}$
So we have that: $\pi_{n}=\Big(\frac{\lambda}{\mu+\delta}\Big)^n\pi_{0}$
However this is a distribution only if $\lambda<\mu+\delta$. And as this has to be true for any $\delta>0$, we must have that $\lambda<\delta$. But this information is not given to me in the problem.
As Mishra Lavrov noted in the comments, the transition rate from state $n$ to state $n-1$ is $\mu+n\delta$. So the detailed balance equations are $$ \lambda\pi_n = (\mu+n\delta)\pi_{n+1}. $$ This yields the recursion $\pi_{n+1}=\frac\lambda{\mu+ n\delta}\pi_n$, with solution $$ \pi_{n+1}=\prod_{k=0}^{n-1}\left(\frac\lambda{\mu+ k\delta}\right)^k\pi_0. $$ Since $\sum_{n=0}^\infty \pi_n=1$, it follows that $$ \pi_0 = \left(1 + \sum_{n=1}^\infty\prod_{k=0}^{n-1}\left(\frac\lambda{\mu+ k\delta}\right)^k\right)^{-1}, $$ and hence $$ \pi_{n+1} = \prod_{k=0}^{n-1}\left(\frac\lambda{\mu+ k\delta}\right)^k\left(1 + \sum_{n=1}^\infty\prod_{k=0}^{n-1}\left(\frac\lambda{\mu+ k\delta}\right)^k\right)^{-1},\; n\geqslant0. $$