I am trying to determine what $Hom_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z})$ is. I believe consists only of the trivial homomorphism. First note that any homomorphism is uniquely determined by its action on $1$. If $f(1) \neq 0$, then I believe you could show that $\Bbb{Q}$ is a cyclic group, which is absurd.
Does this sound right? Here's a sketch. Let $f : \Bbb{Q} \to \Bbb{Z}$ be a $\Bbb{Z}$-module homomorphism, and suppose that $f(1) \neq 0$. Then $f(x) = xf(1)$, and $\Bbb{Q}/ \ker f \cong \Bbb{Z}$, so $\Bbb{Q}/ \ker f$ is cyclic with generator $\overline{q_0}$. Let $x \in \Bbb{Q}$ be arbitrary. Then there exists $n_x \in \Bbb{Z}$ such that $x = n_x \overline{q_0}$ which holds if and only if $x - n_0 q_0 \in \ker f$ if and only if $0 = f(x- n_0 q_0) = (x - n_0 q_0) f(1)$.
Since $f(1) \neq 0$, it follows that $x = n_0 q_0$, which means $\Bbb{Q}$ is cyclic. This is a contradiction, so $f(1) = 0$.
There are two different formulas for the MGF of a Negative Binomial Distribution on Wikipedia, one on the Negative Binomial Distribution page, the other on the MGF page. They don't seem to be algebraically identical. Is one of them wrong?
