Limiting Poisson behaviour in an urn-ball problem This is problem from K.R. Parthasarathy's book  "Introduction to probability and measure."
Consider the placement of r indistinguishable balls in n urns. The sample space here consists of the $\displaystyle {n+r-1 \choose r}$ n-tuples of non-negative integers that add up to r, with the ith entry of the n-tuple denoting the number of balls in urn i. Each point in the sample space is assigned an equal probability of $\displaystyle \dfrac{1}{{n+r-1 \choose r}}.$
Let $p_{n,r}(k)$ denote the probability that exactly $k$ urns are empty.
It can be shown that for each $n$ and $r$ we have
$$
p_{n,r}(k) = \sum_{j = k}^n (-1)^{j-k} { j \choose k} {n \choose j} \frac{ {n-j+r-1 \choose r}  }{{n+r-1 \choose r}}.
$$
For each positive integer $n$, we are given a positive integer $r_n$ such that $\displaystyle \frac{n^2}{r_n} \to \lambda > 0$ as $n \to \infty$.
We are asked to show $\displaystyle p_{n,r_n}(k) \to \exp(-\lambda) \frac{\lambda^k}{k!}$, and I need some help in proving this limiting behavior.
Some Observations
One can see that
$$
p_{n,r}(k) = \frac{1}{k!} \sum_{j = k}^\infty \frac{(-1)^{j-k}}{(j-k)!} q_{n,r}(j),
$$
where $$q_{n,r}(j) = 
\begin{cases} \dfrac{n!}{(n-j)!}\frac{{\genfrac {(} {)} {0pt}{0} {n-j+r-1} {r} }} {{\genfrac {(} {)} {0pt} {0} {n+r-1}{r} }}, \mbox{} \text{ for } 0 \leq j \leq n-1 \\
0, \mbox{} \text{ otherwise}
\end{cases}
$$
So it is sufficient to show that $q_{n,r_n}(j) \leq M^j$ for some $M > 0$ independent of $n$ and $j$ and $q_{n,r_n}(j) \to  \lambda^{j}$ as $n \to \infty$, for then we can take limit inside the summation sign and see that $\lim_{n\to\infty} p_{n,r_n}(j) = \frac{1}{k!} \sum_{j=k}^{\infty} \frac{(-1)^{j-k}}{(j-k)!} \lim_{n \to \infty} q_{n,r_n}(j) = \frac{1}{k!} \sum_{j=k}^{\infty} \frac{(-1)^{j-k}}{(j-k)!} \lambda^j = \exp(-\lambda) \frac{\lambda^k}{k!}.$
 A: \begin{align}q_{n,r}(j)
&= \frac{n!}{(n-j)!}\frac{(n-j+r-1)!}{(n+r-1)!} \frac{(n-1)!}{(n-1-j)!}
\\
&= \frac{[n(n-1) \cdots (n-j+1)][(n-1)(n-2) \cdots (n-j)]}{(n+r-1)(n+r-2) \cdots (n+r-j)}
\\
&= \prod_{m=1}^j \frac{(n+1-m)(n-m)}{n+r-m}
\\
&= \prod_{m=1}^j \frac{(1 + \frac{1-m}{n})(1-\frac{m}{n})}{\frac{r}{n^2} + \frac{n-m}{n^2}}
\\
&\overset{n \to \infty}{\longrightarrow} \prod_{m=1}^j \frac{1}{1/\lambda} 
\\
&= \lambda^j.
\end{align}
A: We have, for a fixed $j$,
$$
\begin{align}
q_{n,r}(j) &= \frac{n!}{(n-j)!} \frac{ n-j+r-1 \choose r}{n+r-1 \choose r}\\
&=  \left\{ n (n-1) \dots (n-j+1) \right\}\left\{\frac{ (n-j) (n-j+1) \dots (n-j+r-1) }{ (n)(n+1)\dots (n+r-1)} \right\}\\
& = \prod_{t = 0}^{j-1} (1 - \frac{t}{n}) \times n^j \times \prod_{t=0}^{r-1}(1-\frac{j}{n+t})\tag{1}
\label{e:1}
\end{align}
$$
We will show the third term in the above product satisfies $\displaystyle \prod_{t=0}^{r-1}(1-\frac{j}{n+t}) \leq  \left(\frac{n}{n+r}\right)^j.$
Proof:

From the identity $1-x \leq \exp(-x)$ we have $\displaystyle 0 \leq \prod_{t=0}^{r-1}(1-\frac{j}{n+t}) \leq \exp( -j \sum_{t=0}^{r-1}\frac{1}{n+t})$
For any $p > 0$ we have $\displaystyle \frac{1}{p} \geq \int_{p}^{p+1}\frac{1}{x}dx \geq \log \frac{p+1}{p}$.
This implies $\displaystyle \sum_{t=0}^{r-1}\frac{1}{n+t} \geq \sum_{t=0}^{r-1} \log \frac{n+t+1}{n+t} = \log \frac{ n+ r}{n}$ and hence $\displaystyle\exp(-j \sum_{t=0}^{r-1}\frac{1}{n+t}) \leq \exp(-j \log\frac{n+r}{n} ) \leq \left(\frac{n}{n+r}\right)^j$.

The above inequality implies, on subsitution in $\eqref{e:1}$ $\displaystyle 0 \leq q_{n,r}(j) \leq n^j \left(\frac{n}{n+r}\right)^j = \left( \frac{n^2}{n+r} \right)^j = \left(\frac{ \frac{n^2}{r} } {\frac{n}{r} + 1}\right)^j$.
So, $\displaystyle 0 \leq q_{n,r_n}(j) \leq \lambda_n^j $ where $\displaystyle \lambda_n = \frac{n^2/r_n}{n/r_n+1} \to \lambda$, in particular $\displaystyle q_{n,r_n}(j) \leq M^j$ for some suitable $M > 0$ and $\limsup_{n\to\infty} q_{n,r_n}(j) \leq \lambda^j \tag{2a} \label{e:2a}.$
So it is sufficient to prove $q_{n,r_j}(j) \to \lambda^j.$
We will now show that for all sufficiently large $n$, $$ \displaystyle \prod_{t=0}^{r-1}(1-\frac{j}{n+t}) \geq \left(\frac{n-1}{n+r-1}\right)^j \exp(-j^2\sum_{t=n}^{\infty}\frac{1}{t^2}) \tag{2} \label{e:2}.$$
We may assume $n$ is so large that $\frac{j}{n} \leq \frac{1}{2}$.
For any $0 \leq x \leq \frac{1}{2}$ we have $(1-x) \geq \exp(-x-x^2)$. This follows from exponentiating both sides of the following inequality: $-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots \leq x+\frac{x^2}{2}(1+x+x^2 + \dots + ) \leq x + \frac{x^2}{2}(1+\frac{1}{2}+\frac{1}{2^2}+\dots) = x+x^2.$
Hence, $$
\begin{align}
\displaystyle  \prod_{t=0}^{r-1}(1-\frac{j}{n+t}) &\geq \exp\left( -j \sum_{t=0}^{r-1}\frac{1}{n+t} - j^2\sum_{t=0}^{r-1}\frac{1}{(n+t)^2}\right)\\
& \geq \exp\left( -j \sum_{t=0}^{r-1}\frac{1}{n+t} - j^2\sum_{t=n}^{\infty}\frac{1}{t^2}\right). \tag{3}
\label{e:3}
\end{align}
$$.
For any $p > 1$ we have $\displaystyle \frac{1}{p} \leq \int_{p-1}^{p}\frac{1}{x}dx \leq \log \frac{p}{p-1}$. Hence $\displaystyle \sum_{t=0}^{r-1}\frac{1}{n+t} \leq \sum_{t=0}^{r-1} \log \frac{n+t}{n+t-1} \leq \log\frac{n+r-1}{n-1}$, and $\eqref{e:2} $ follows from substituting this inequality in $\eqref{e:3}$.
So from $\eqref{e:1}$ and $\eqref{e:2}$ we have, $$
\begin{align}
q_{n,r}(j) &\geq \left\{\prod_{t=0}^{j-1}(1-\frac{t}{n})\right\} n^j \left( \frac{n-1}{n+r-1}\right)^j \exp(-j^2\sum_{t=n}^{\infty} \frac{1}{t^2}) \\
&\geq \left\{\prod_{t=0}^{j-1}(1-\frac{t}{n})\right\} \left( \frac{\frac{n^2}{r} -\frac{n}{r}}{\frac{n}{r}+1-\frac{1}{r}}\right)^j \exp(-j^2\sum_{t=n}^{\infty} \frac{1}{t^2}).
\end{align}
$$
So, putting $r = r_n$ above and letting $n \to \infty$ we have,
$\liminf_{n \to \infty} q_{n,r_n}(j) \geq \lambda^j \tag{4} \label{e:4}.$
From $\eqref{e:2a}$ and $\eqref{e:4}$ we have $\lim_{n \to \infty} q_{n,r_n}(j) = \lambda^j$ and the result follows.
