Following situation: I look at a queueing system with Poisson arrivals with rate $\lambda=4/15$, each arriving customer has to go through r=2 phases. In each phase, the service time is exponential distributed with rate $\mu =1$. Customers are served in order of arrival.
I should calculate the mean number of jobs in the system. I already computed the distribution of the number of uncompleted phases $$ p_n=\frac{7}{24} \Big( \frac{2}{3}\Big)^n+\frac{7}{40} \Big(-\frac{2}{5} \Big)^n $$ as well as the distribution of the number of customers in the system: $$ q_n=\frac{35}{48} \Big( \frac{4}{9}\Big)^n-\frac{21}{80} \Big(-\frac{4}{25} \Big)^n $$ where I used for the latter that $$q_i=\sum_{n=(i-1)r+1}^{ir} p_n$$.
My problem is to compute the mean number of customers in the system. In the solution to this exercise, it just says $$\mathbb{E}[L]=\sum_{n=0}^{\infty} n q_n=104/105$$. Putting my solution for $q_n$ into Wolfram Alpha gives exactly this result, but how do I actually compute this "by hand"?
(This question is about exercise 30 resp. 33 of the book "Queueing Systems" from Ivo Adan and Jacques Resing.)
The analytical approach is to use the fact that $$\sum_{n=1}^\infty nr^{n-1} = \frac{1}{(1-r)^2}.$$ You'll have to write $q_n$ in the form of $r^{n-1}$ to do this.
The numerical approach is just to calculate the first $N$ terms of the sum, for some $N$. After some point, the terms $nq_n$ are going to get very, very small, and you can start to ignore them.