p(y)=$\frac{\Gamma(y+\kappa)}{y!\Gamma(\kappa))}\frac{\lambda^\kappa}{(1+\lambda)^{y+\kappa}}$
I want to show that as $\kappa->\infty$ with $\mu=\kappa/\lambda$ held constant that p(y) converges to the probability mass funciton of the Poisson($\mu$).
But what does $\mu=\kappa/\lambda$ held constant mean? Does it mean that as $\kappa->\infty$ that $\lambda$ will also increase so that the ratio $\kappa/\lambda$ will always be constant?
Otherwise, as $\kappa->infty$ $\mu$ must also increase right?
Yes, saying $\mu=\frac \kappa \lambda$ is constant means $\lambda$ increases too. You can just substitute $\frac \kappa \mu$ for $\lambda$, then let $\kappa \to \infty$