Suppose that $\pi(\cdot,\cdot,\cdot)$ is a probability distribution on $\{0,...,N\}^3$ with $$ \pi(i,j,k) \propto \frac1{\nu^i} \frac1{\lambda^j} \frac1{\mu^k} \frac1{k!}. $$ I then wish to show that $$ p = \sum_{j+k=N} \pi(0,j,k) = \frac{H(N)}{\sum_{n=0}^N H(n)} \quad \text{where} \quad H(n) = \biggl(\frac\nu\lambda\biggr)^n \sum_{\ell=0}^n \biggl(\frac\lambda\mu\biggr)^\ell \frac1{\ell!}.$$
To this end, I wrote $$ \pi(i,j,k) = G^{-1} \frac1{\nu^i} \frac1{\lambda^j} \frac1{\mu^k} \frac1{k!} \quad \text{where} \quad G = \sum_{i+j+k=N} \Bigl( \nu^i \lambda^j \mu^k k! \Bigr)^{-1}.$$ We then define $$ f(r) = \sum_{j+k=r} \Bigl( \lambda^j \mu^k k! \Bigr)^{-1} \quad \text{and so} \quad G = \sum_{i=0}^N \nu^{-i} f(N-i) = \sum_{i=0}^N \nu^{i-N} f(i).$$ Finally, we let $g(i) = \nu^i f(i)$ to obtain $$ p = \frac{g(N)}{\sum_{i=0}^N g(i)}. $$ Straightforward calculation shows that $$ g(r) = \nu^r \sum_{j+k=r} \Bigl(\lambda^{r-k} \mu^k k! \Bigr)^{-1} = \biggl(\frac\nu\lambda\biggr)^r \sum_{k=0}^r \sum_{j=0}^{r-k} \biggl(\frac\lambda\mu\biggr)^k \frac1{k!}. $$ This is almost exactly as required, except for the pesky $\sum_{j=0}^{r-k}$. Note that there is no $j$ term in the summand, so this just gives a multiplicative factor of $(r+1-k)$, ie $$ g(r) = \biggl(\frac\nu\lambda\biggr)^r \sum_{k=0}^r \bigl(r+1-k\bigr) \biggl(\frac\lambda\mu\biggr)^k \frac1{k!}. $$
This is as far as I have been able to get. Any assistance for completion would be most appreciated!
This exercise is taken from Stochastic Networks Book (Kelly) Exercise 2.6.