Given the Erlang B Formula: $$P_n = \frac{\dfrac{\rho^n}{n!}}{\sum_{l=0}^{N} \dfrac{\rho^l}{l!}}$$ This exists for $n = 0,1,\ldots,N$ and for $n>N$, $P_n = 0$.
According to my textbook, the expected value is $$E(n) = (1-P_N)\rho$$ but I don't understand how to get here.
I start by setting the two sides equal to each other: $$ E(n) = (1-P_N)\rho = \sum_{k=0}^{N} kp_k. $$ Then, I can recognize that $$(1-P_N) = \sum_{l=0}^{N-1} \rho^l$$ since it's functionally the probability of every state but N. After that I'm not sure where to go.
Let $1/c = \sum_{l=0}^{N}\frac{\rho^l}{l!}$.
Then, $E(n) = c \sum_{n=0}^{N} n P_n = c \sum_{n=0}^{N} n \rho ^n/n! = = c \rho \sum_{n=1}^{N} \rho ^{n-1}/(n-1)!$.
Substituting $k = n-1$, $E(n) = c \rho \sum_{k=0}^{N-1} \rho ^{k}/k! = \rho (1 - P_N)$.