The Negative Binomial RV $X$ models the number of trials until the $r$-th success in a sequence of independent Bernoulli Trials with probability of success $p$ in each trial. So, if $q = 1 - p$,
$$P(X = k)= \begin{pmatrix} k-1 \\r-1 \end{pmatrix}p^rq^{k-r},k=r,r+1,\dots.$$
Let $Y = X - r$.
Suppose that $r\to\infty$ and $q\to 0$ so that $rq\to\lambda$.
Show that, for fixed $m=0,1,\dots$
$$P(Y=m)\longrightarrow\frac{\lambda^m}{m!}e^{-\lambda}.$$
So what I did so far was
Let $r=\frac{\lambda}{q}$,with $q\to 0$
Then $P(Y=m) = P(X=r+m)$
So I got
$$P(X = r+m)=\begin{pmatrix}\frac{\lambda}{q}+m-1\\ \frac{\lambda}{q}-1 \end{pmatrix} p^{\frac{\lambda}{q}}q^{m}, m=0,1,\dots.$$
and I don't know what to do next, can someone help me out ?
Take the limit as $q\rightarrow 0,$ but express the last line as $$P(X=r+m)=\frac{(\frac{\lambda }{q}+m-1)!}{m!(\frac{\lambda }{q}-1)!}(1+\frac{1}{-1/q})^{-\lambda /q}q^m\frac{r^m}{r^m}=\underbrace{\frac{(\frac{\lambda }{q}+m-1)!}{r^m(\frac{\lambda }{q}-1)!}}_{\rightarrow 1}\underbrace{(1+\frac{1}{-1/q})^{-\lambda /q}}_{\rightarrow \color{red}{e^{-\lambda}}}\color{red}{\frac{\lambda^m}{m!}}$$ use that $rq=\lambda ,$ use the known limit $(1+1/x)^x\longrightarrow e$ as $x\rightarrow \infty$ and use that $$\frac{(r+m-1)!}{r^m(r-1)!}=\frac{r^{\overline{m}}}{r^m}\longrightarrow 1,$$ the las line i would use the expansion in Stirling numbers of the raising factorial but i guess is just an argument of the degree of the polnomial in $r.$