Negative Binomial MGF converges to Poisson MGF This question is Exercise 3.15 in Statistical Inference by Casella and Berger. It asks to prove that the MGF of a Negative Binomial $\mathscr{N}eg(r,p)$ converges to the MGF of a Poisson $\mathscr{P}(\lambda)$ distribution, when 
$$r\to\infty\,, \quad p\to 1\,, \quad r(1-p)\to\lambda$$
The formula I have for the MGF of $X\sim \mathscr{N}eg(r,p)$ is:
$$M_X(t) = \frac{p^r}{[1-e^t(1-p)]^r}$$
Considering just the denominator, we have 
$$[1-e^t(1-p)]^r = [1+\frac{1}{r}e^tr(p-1)]^r = [1+\frac{1}{r}e^t(-\lambda)]^r$$ As $r\to\infty$, this converges to $e^{-\lambda e^t}$. Now considering the entire formula again, and letting $r\to\infty$ and $p\to 1$, we get $e^{\lambda e^t}$, which is incorrect since the MGF of Poisson($\lambda$) is $e^{\lambda(e^t-1)}$. I seem to be on the right track, just made a misstep somewhere. Can anyone spot my mistake?
 A: You make a mistake by ignoring $p^r$: If you consider your MGF $$M_X(t) = \frac{p^r}{[1-e^t(1-p)]^r}\,,$$
then $$\log\{M_X(t)\} = r\log(p)-r\log\{1-e^t(1-p)\}$$ and using the asymptotic equivalences
\begin{align*}
r\log(p)-r\log\{1-e^t(1-p)\}
&= r\log(1-[1-p])-r\log\{1-e^t(1-p)\}\\
&\approx -r[1-p]+re^t(1-p)\\
&\approx \lambda[-1+e^t]
\end{align*}
which shows that the limiting value of the MGF is
$$ \exp\{\lambda[e^t-1]\}$$ as requested in this exercise.

Note: There are two versions of the MGF for a eg(n,p) distribution,
  one for the number of trials and one for the number of failures. The
  current version is the MGF for the number of failures, which starts at
  zero like the Poisson distribution.

A: $$\eqalign{    M_{x}(t)&=\left[\frac{p}{1-\frac{e^{t}(1-p)}{p}}\right]^r \\
             &=\left[\frac{1-(1-p)}{1-\frac{e^{t}(1-p)}{p}}\right]\\
             &= \left[\frac{1-\frac{r(1-p)}{r}}{1-\frac{e^{t}r(1-p)}{rp}}\right]\\
             &=\frac{e^{-r(1-p)}}{e^{{-e^{t}}r(1-p)}}\\
             &=e^{-\lambda}e^{e^{t}\lambda}
}$$
by applying the given conditions, and we have the desired result
A: The above answer was correct that you ignored the numerator $p^r$, but I found the illustration is a little confusing, so here we go for a standard way to solve the problem
$\begin{aligned} \lim_{r\to \infty} M_{NB}(t) & = \lim_{r\to \infty} (\frac{p}{1-(1-p)e^t})^r \\ &= \lim_{r\to \infty} (\frac{1-(1-p)}{1-(1-p)e^t})^r \\ &= \lim_{r\to \infty} \frac{(1 + \frac{1}{r}(-\lambda))^r }{( 1+ \frac{1}{r}e^t(-\lambda))^r } \quad \quad r(1-p) =  \lambda \Rightarrow 1-p = \frac{\lambda}{r} \\ &=  \frac{e^{-\lambda}}{e^{-\lambda e^t}} \\ &= e^{\lambda(e^t-1)}  \end{aligned}$
which is the $MGF$ of Poisson Distribution
Note: Lemma 2.3.14 in the Statistical Inference by Casella and Berger book states a useful limit where when we have $\lim_{n\to\infty} a_n = a$
$\lim_{n\to\infty} (1 + \frac{a_n}{n})^n = e^a$
