Error Bound in Binomial Approximation to Poisson In formula $(1.1)$ on page $1$ of the book Poisson Approximation by Barbour, Holst, and Janson, they assert the formula
$${n \choose k} p^k (1-p)^{n-k} = \frac{(np)^k}{k!} e^{-np} \left(1 + \mathcal{O}(np^2, k^2 n^{-1}) \right).$$
Here $0 \leq k \leq n,$ and $p \in [0,1].$ They say that this can be derived by an "elementary but somewhat involved calculation". How do we derive it? As a first step, how is $\mathcal{O}(a,b)$ defined?
I can rearrange the question into a more usual "big O" problem: we want to show
$$\frac{e^{np} (1-p)^{n-k} n^{\underline{k}}}{n^k} - 1 = \mathcal{O}(np^2, k^2 n^{-1}).$$
Here $n^{\underline{k}} := n (n-1) \cdots (n-k+1).$
I have tried expanding out the expression on the left hand side- the fraction can be written as the product
$$\left(\sum_{j=0}^\infty \frac{(np)^j}{j!}\right) \left(\sum_{i=0}^{n-k} (-1)^i p^i {n-k \choose i}\right) \left(\sum_{l=0}^k s(k,k-l) n^{-l} \right),$$
Where $s(k, l)$ is a Stirling number of the first kind. At this point, I don't know how to proceed.
 A: My favorite is generating functions (G F) approach.
G F of the binomial distribution
$$P(s)=\sum_{k=0}^{n}{n \choose k}(1-p)^{n-k}p^k s^k=(1-p+ps)^n$$
Let's introduce the notation $\lambda= np$
$$P(s)=\left (1 -\lambda\frac{1-s}{n} \right )^n$$
Now,
$$P(s)\rightarrow e^{-\lambda (1-s)}=e^{-\lambda}e^{\lambda s} $$
as $n\rightarrow \infty$
So we came to the G F of the poisson distribution.
Improvements
Actually, I don't know what $\mathcal{O}(a,b)$ is but as a physicist i can suggest the following elementary procedures to estimate the precision of the expression
$$\frac{{n \choose k}(1-p)^{n-k}p^k }{e^{-\lambda}\frac{\lambda^k}{k!}}$$
where $\lambda= np$
After taking the logarithmic derivative of the expression we get
$$\ln( n-k )+\frac{1}{2( n-k )}+\ln\frac{p}{\lambda (1-p)}$$
(Stirling approximation $n!=\sqrt{2\pi n}\left (\frac{n}{e}\right )^n$ was used.)
Equating the result to zero and solving for k we get
$$k=np=\lambda$$
(Unimportant term $\frac{1}{2( n-k )}$ was ignored)
At $k=np=\lambda$ the initial expression has maximum equal to
$$\frac{e^{\lambda} \lambda! }{\sqrt{2\pi \lambda}\lambda^{\lambda}}=1+\frac{1}{12\lambda}+\frac{1}{288\lambda^2}+...$$
Here we used the correction terms of the Stirling approximation.
Finally, we get the following estimate
$$\frac{{n \choose k}(1-p)^{n-k}p^k }{e^{-\lambda}\frac{\lambda^k}{k!}}<1+\frac{1}{12\lambda}+\frac{1}{288\lambda^2}+...$$
$$1\leq \lambda$$
