# Continuous-Service M/M/1 Queuing Systems
## Appendix A  
Given the steady-state distribution \(\pi=\{\ldots,\pi_{-2},\pi_{-1},\pi_{0},\pi_{1},\pi_{2},\ldots\}\) of our continuous-service M/M/1 queueing systems and \(0\leq\rho<1\), \(\sum_{j=0}^{\infty}\Bigl{(}\pi_{-j}+\pi_{j+1}\Bigr{)}=\rho^{n}\) for \(n=0,\ 1,\ 2,\ \ldots\). We prove the above identity by induction. We first verify that the above identity holds for \(n=0\). In fact the identity follows in that, when \(n=0\), we have \(\sum_{j=0}^{\infty}\Bigl{(}\pi_{-j}+\pi_{j+1}\Bigr{)}=(\pi_{0}+\pi_{1})+(\pi_ {-1}+\pi_{2})+(\pi_{-2}+\pi_{3})+\cdots=1=\rho^{0}\) since all probabilities sum to \(1\). We next assume that the identity holds for some positive integer \(\mathsf{k}\). That is, we have \(\sum_{j=k}^{\infty}\Bigl{(}\pi_{-j}+\pi_{j+1}\Bigr{)}=(\pi_{-k}+\pi_{k+1})+( \pi_{-k-1}+\pi_{k+2})+(\pi_{-k-2}+\pi_{k+3})+\cdots=\rho^{k}\). We finally need to show that the identity holds for \(k+1\). That is, we must show that \(\sum_{j=k+1}^{\infty}\Bigl{(}\pi_{-j}+\pi_{j+1}\Bigr{)}=(\pi_{-k-1}+\pi_{k+2})+ (\pi_{-k-2}+\pi_{k+3})+\cdots=\rho^{k+1}\). To this end, we see that  
\[\begin{array}{ll}\sum_{j=k+1}^{\infty}\Bigl{(}\pi_{-j}+\pi_{j+1}\Bigr{)}&=( \pi_{-k-1}+\pi_{k+2})+(\pi_{-k-2}+\pi_{k+3})+\cdots\\ &=(\pi_{-k}+\pi_{k+1})+(\pi_{-k-1}+\pi_{k+2})+(\pi_{-k-2}+\pi_{k+3})+\cdots-( \pi_{-k}+\pi_{k+1})\\ &=\rho^{k}-(\pi_{-k}+\pi_{k+1})\\ &=\rho^{k}-\frac{\rho^{k}(1-\rho)}{(1+\rho)^{k+1}}-\rho^{k}(1-\rho)\Bigl{[}1- \frac{1}{(1+\rho)^{k+1}}\Bigr{]}\,(\text{Use Equation }(\ref{eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq: eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq:eq: eq:  
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