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

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?

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.

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

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 $$R \to 1$$, $$0 \le \frac{\lambda^\nu}{\nu!} -S_{\nu} \to 0$$