I need to show that the Poisson probability distribution, given by $P(n|\mu)=\frac{\mu^{n}e^{-\mu}}{n!}$, is a well-behaved probability, i.e. $\sum^{\infty}_{n=0}P(n|\mu)=1$ and $\int_0^{\infty} P(n|\mu)d\mu =1$.
I have done the first part by comparing it to the expansion of $e^\mu$, but am unsure how to do the second part. When I take the integral I end up with terms of $\infty^n$ and I don't know what to do with those.