My goal is to show that in a $M/M/c$ queueing system it is satisfied that
$$ L_s = L_q + \frac{\lambda}{\mu}, $$
where $L_s$ represents the average number of costumers in the system, $L_q$ represents the average number of costumers in the queue and $\lambda$,$\mu$ represent the usual arrival rate and service rate, respectively.
Prior to this, I was able to show that $L_q = P_c\frac{\lambda/(c\mu)}{(1-\lambda/(c\mu))^2}$, where $P_c$ represents the probabily that there are $c$ costumers in the system, when the system is in steady-state. Altought, I don't think this is going to be usefull to the proof I am now trying to do.
We have that
$$ L_s = \sum_{n=0}^\infty n P_n = \sum_{n=0}^{c} nP_n + \sum_{n=c+1}^\infty (n-c+c)P_n = \sum_{n=0}^c n P_n + c \sum_{n = c+1}^\infty P_n + \sum_{n=c+1}^\infty (n-c)P_n. $$
Now, by definition we have that the last term equals $L_q$, and it is easy to evalue the second term:
$$ \sum_{n=c+1}^\infty P_n = \sum_{n=c+1}^\infty \left( \frac{\lambda}{\mu} \right)^n \frac{1}{c!c^{n-c}}P_0 = \frac{P_0\, c^c}{c!} \sum_{n=c+1}^\infty \left( \frac{\lambda}{c\mu} \right)^n = \frac{P_0 \, c^c}{c!} \, \frac{\left( \frac{\lambda}{c\mu} \right)^{c+1}}{1-\frac{\lambda}{c\mu}}. $$
Hence, we have that
$$ L_s = \sum_{n=0}^c nP_n + c \frac{P_0 \, c^c}{c!} \, \frac{\left( \frac{\lambda}{c\mu} \right)^{c+1}}{1-\frac{\lambda}{c\mu}} + L_q = \sum_{n=0}^c nP_n + \frac{P_0 \left( \frac{\lambda}{\mu} \right)^{c+1}}{c!(1-\frac{\lambda}{c\mu})} + L_q.$$
Please note that $ P_0 = \left( \sum_{n=0}^{c-1} \left( \frac{\lambda}{\mu} \right)^n \frac{1}{n!} + \frac{(\lambda/\mu)^c}{c!(1-\frac{\lambda}{c\mu})} \right)^{-1}.$ So now all I need to do is to show that
$$ \sum_{n=0}^c n P_n + \frac{P_0 \left( \frac{\lambda}{\mu} \right)^{c+1}}{c!(1-\frac{\lambda}{c\mu})} = \frac{\lambda}{\mu}.$$
This is where I am having difficulties. We have that
$$ \sum_{n=0}^c nP_n = \sum_{n=0}^c n \left( \frac{\lambda}{\mu} \right)^n \frac{P_0}{n!} = P_0 \sum_{n=1}^c \left( \frac{\lambda}{\mu} \right)^n \frac{1}{(n-1)!}. $$
But I don't know how to proceed after this. Perhaps there is a simpler approach that I am not recognizing?
Thanks for any help in advance.
I'm going to write a proof that $$ \sum_{n=0}^c n P_n + \frac{P_0 \left( \frac{\lambda}{\mu} \right)^{c+1}}{c!(1-\frac{\lambda}{c\mu})} = \frac{\lambda}{\mu}$$
Proof :
Since $\displaystyle\sum_{n=1}^{c}\left(\frac{\lambda}{\mu}\right)^n\frac{1}{(n-1)!}$ can be written as $\displaystyle\sum_{n=0}^{c-1}\left(\frac{\lambda}{\mu}\right)^{n+1}\frac{1}{n!}$, we have $$\begin{align}&\sum_{n=0}^c n P_n + \frac{P_0 \left( \frac{\lambda}{\mu} \right)^{c+1}}{c!(1-\frac{\lambda}{c\mu})} \\\\&= P_0 \sum_{n=1}^c \left( \frac{\lambda}{\mu} \right)^n \frac{1}{(n-1)!}+\frac{P_0 \left( \frac{\lambda}{\mu} \right)^{c+1}}{c!(1-\frac{\lambda}{c\mu})} \\\\&=P_0 \sum_{n=0}^{c-1}\left(\frac{\lambda}{\mu}\right)^{n+1}\frac{1}{n!}+\frac{P_0 \left( \frac{\lambda}{\mu} \right)^{c+1}}{c!(1-\frac{\lambda}{c\mu})} \\\\&=\frac{\lambda}{\mu}\times\underbrace{P_0\bigg(\sum_{n=0}^{c-1}\left(\frac{\lambda}{\mu}\right)^{n}\frac{1}{n!}+\frac{\left( \frac{\lambda}{\mu} \right)^{c}}{c!(1-\frac{\lambda}{c\mu})}\bigg)}_{=1} \\\\&=\frac{\lambda}{\mu}\end{align}$$