# Separate limits in binomial coefficient series

In reading an old book I stumbled upon this expression

$$\sum_{k=0}^n {n \choose k} q^k (1-q)^{n-k} f(n,k)=0$$

With the trick of setting $\lambda=nq$ the author lets $n \to \infty$ and $q \to 0$, resulting in

$$\sum_{k=0}^{\infty} e^{-\lambda} \frac{\lambda^k}{k!} \lim_{n,q \to \infty} f(n,k)=0.$$

The limit of the summand is taken and n is summed up to infinity. But can those limits be taken separately? Is there a proper way to derive the expression given above?

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Which book is this? – Did Sep 2 '12 at 18:12
What if you simply replace $q$ with $\lambda\over n$ and let $n\to \infty$? – Hagen von Eitzen Sep 2 '12 at 18:52
The function $f(n,k)$ is quite specific since $f(n,k)=u(a_n-k)$ for some $a_n\to a$ and, presumably, $u$ continuous. Thus your sum is $E(u(a_n-X_n))$ where $X_n$ is binomial $(n,q)$, hence $X_n$ converges in distribution to $X$ Poisson $\lambda$, and the limit is $E(u(a-X))$. – Did Sep 2 '12 at 19:08
You showed the source in a comment, then deleted the comment. Why? – Did Sep 9 '12 at 15:10

The function $f(n,k)$ is quite specific since $f(n,k)=u(a_n-k)$ for some $a_n\to a$ and, presumably, $u$ continuous. Thus your sum is $E(u(a_n-X_n))$ where $X_n$ is binomial $(n,q)$, hence $X_n$ converges in distribution to $X$ Poisson $\lambda$, and the limit is $E(u(a-X))$.