I have a function like the following:
$$f(u) = \sum^u_{v=0} \frac{e^{-\theta}\theta^{u-v}}{(u-v)!}*\frac{e^{-\lambda}\lambda^v}{v!} $$
I am trying to get the answer of $$\frac{e^{-(\theta + \lambda)}(\theta + \lambda)}{u!},$$ but I am unsure how to get to this expression. How can I simplify $f(u)$? Thanks!
$$f(u)=\sum_{v=0}^{u} \frac{e^{-t}t^{u-v}}{(u-v)!} ~ \frac{e^{-s} s^v}{v!}=\frac{e^{-t-s} t^u}{u!} \sum_{v=0}^{u}\frac{u! (s/t)^v}{v!(u-v)!}=\frac{t^ue^{-(s-t)}}{u!} \sum_{v=0}^{u} {u \choose v} (s/t)^v.$$ $$\implies f(u)=\frac{t^ue^{-(s+t)}}{u!} (1+s/t)^u=\frac{e^{-(s+t)}}{u!} (s+t)^u$$