Probability question that links between $NB$ and $Gamma$ distributions with limits. 
Let $\lambda>0$, and $ m\ge 1$, $m\in\mathbb{N}$. 
For every $n\ge 1:$ $Y\sim NB(m, \frac{\lambda}{n})$, we define: $\hat Y_n=\frac{1}{n}Y_n$. Show that for every $t\ge 0$: 
$F_{\hat Y_n}(t)\to F_{\hat Y}(t)$ when $n\to \infty$ 
While $\hat Y\sim Gamma(m, \lambda)$ or in other words: 
$f_{\hat Y}(y)=\frac{1}{(m-1)!}\lambda^my^{m-1}e^{-\lambda y}$. whenever $y\ge 0$, and equal $0 $ otherwise. (Would be happy to learn how to write the function with the otherwise in latex).

My Attempt:
$F_{\hat Y_n}(t)=P(\hat Y_n\le t)=P(Y_n \le nt)=\sum^{[nt]}_{k=m} \binom {k-1}{m-1}(\frac{\lambda}{n})^{m}(1-(\frac{\lambda}{n})^{k-m})=(\frac{\lambda}{n})^{m}\sum^{[nt]}_{k=m}\binom {k-1}{m-1}(1-(\frac{\lambda}{n})^{k-m})$ 
I'm currently stuck with this sum, I'm not sure what should be my next step, I'm just trying to find $F_{\hat Y_n}(t)$, and take the limit when $n\to\infty$ 
Any help is really appreciated, thanks in advance!
 A: Note that $\lim_{n\rightarrow\infty}F_{\hat Y_n}= F_{\hat Y}$ is the same as $\lim_{n\rightarrow\infty}f_{\hat Y_n}=f_{\hat Y}$. It is easier to work with the pdf than the cdf so we will do so.
Let $X=\frac 1n Y_n$. Then the transformation theorem says that the inverse function $s(x)=y_n=nx$ and $\frac{ds(x)}{dx}=n$, so $g(x)=nf(nx)=n{m+nx-1\choose nx}(\frac \lambda n)^m(1-\frac \lambda n)^{nx}$ and $0\le x<\infty$.
Now take the limit in $n$. We know that $(1+a_n)^{c_n}$ converges to $e^{a_nc_n}$ if $a_n\rightarrow 0$ and $a_n^2c_n\rightarrow 0$. Thus $(1-\frac \lambda n)^{nx}\rightarrow e^{-\lambda x}$. Now examine $$\begin{split}\lim_{n\rightarrow \infty} \frac{(m+nx-1)!}{(nx)!n^{m-1}}&=\lim_{n\rightarrow\infty} \frac{(nx+m-1)\cdot(nx+m-2)\cdot\dots\cdot(nx+1)}{n\cdot n\cdot \dots\cdot n \text{ [$m-1$ times}]}\\
&=x^{m-1}\end{split}$$
Thus we get
$$\lim_{n\rightarrow\infty} g(x)=\frac{\lambda^m}{(m-1)!}x^{m-1}e^{-\lambda x}$$
This is the pdf of a $\text{Gamma}(m, \lambda)$ distribution. Thus the asymptotic distribution of $\hat Y_n$ is $\hat Y$.
