classical occupancy problem in Feller - $r$ balls in $n$ cells - poisson approximation

Question

I'm reading Feller's Introduction to Probability Vol. 1 page 103, and I'm try to wrap my head around the following.

There is a step from the section on classical occupancy problem (section 2 of chapter IV). The classical occupancy problem involves the random distribution of $r$ balls in $n$ cells, where we seek the probability $p_m(r,n)$ of finding exactly $m$ cells empty.

In Chapter IV, (2.6) we have

(2.6) $\hspace{1in}$ $ \{ne^{-(\nu+r)/(n-\nu)} \}^\nu < \nu! S_{\nu} < \{ne^{-r/n}\}^\nu$

Then $\lambda$ is set $$ne^{-r/n} = \lambda $$

and suppose that $r$ and $n$ increase in such a way that $\lambda$ remains constrained to a finite interval: $0 < a < \lambda < b$.

For each fixed $\nu$ the ratio of extreme members in (2.6) then tend to unity, and so

$$ 0 \leq \frac{\lambda^\nu}{\nu!} - S_{\nu} \to 0 $$

How did he arrive at the last step?

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Appendix: $S_\nu$ is defined as below $$ S_\nu = {n \choose \nu} \left( 1- \frac{\nu}{n} \right)^r$$ for every $ \nu \leq n $

For example,

$ S_1 = \Sigma p_i $, where $p_i$ is the probability that the $i$th bin is empty.

$ S_2 = \Sigma p_{ij} $, where $p_{ij}$ is the probability that the $i$th and $j$th bins are empty, for all $i$ and $j$ and $i<j$.

$ S_3 = \Sigma p_{ijk} $, ... and so on.

Answer 1

Define $$R = \left( \frac{n \exp[-(\nu+r)/(n-\nu)]}{n \exp(-r/n)} \right)^\nu$$ The book claims, and it's not too hard to show, that $R \to 1$ as $r, n \to \infty$. Since the posted question is about "the last step", we will leave a proof of this first part to the reader. With the above definition of $R$, the initial inequality can then be written $$ R \cdot (n \exp(-r/n))^\nu < \nu! S_{\nu} < (n \exp(-r/n))^\nu$$ With the substitution $\lambda = n \exp(-r/n)$, $$ R \cdot \lambda^\nu < \nu! S_{\nu} < \lambda^\nu$$ so $$ R \cdot \frac{\lambda^\nu}{\nu!} < S_{\nu} < \frac{\lambda^\nu}{\nu!}$$ $$ - \frac{\lambda^\nu}{\nu!} < -S_{\nu} < - R \cdot \frac{\lambda^\nu}{\nu!}$$ $$ 0 < \frac{\lambda^\nu}{\nu!} -S_{\nu} < (1-R) \cdot \frac{\lambda^\nu}{\nu!}$$ From this last inequality, we see that as $R \to 1$, $$0 \le \frac{\lambda^\nu}{\nu!} -S_{\nu} \to 0$$