Let $f:\mathbb R\to\mathbb R$ be continuous and bounded. Prove that for each $x>0$ we have
$$f(x)=\lim_{n\to\infty}\left(e^{-nx}\cdot\sum_{k=0}^\infty\frac{(nx)^k}{k!}f\left(\frac{k}{n}\right)\right).$$
When $x$ is an integer we have
$$e^{-nx}\cdot\sum_{k=0}^\infty\frac{(nx)^k}{k!}f\left(\frac{k}{n}\right)=\frac{e^{-nx}}{(nx)!}\cdot\sum_{k=0}^\infty\binom{nx}{k}(nx-k)!(nx)^k.$$
I then substitute the gamma function integral formula for the factorial, swap the order of $\sum$ and $\int$ and apply the binomial theorem to get that this is equal to
$$\frac{e^{-nx}}{(nx)!}\int_0^\infty(nx+y)^{nx}e^{-y}\,dy.$$
This doesn't look easy to solve. When $x$ is not an integer, I am having even more difficulties.
I would appreciate any help.
Best Answer
Here's a fun probabilistic proof.
Let $X_1,X_2,\dots$ be iid Poisson random variables with parameter $x>0$. Then $S_n=X_1+\dots+X_n$ is Poisson with parameter $nx$.
By the weak law of large numbers, $S_n/n\stackrel{\mathbb P}\to\mathbb EX_1=x$, so $S_n/n\stackrel{d}\to x$. Hence for any continuous bounded $f$, we have $\mathbb E[f(S_n/n)]\to f(x)$ as $n\to\infty$, i.e. $$f(x)=\lim_{n\to\infty}\left(\sum_{k=0}^\infty f(k/n)\cdot\mathbb P(S_n=k)\right)=\lim_{n\to\infty}\left(\sum_{k=0}^\infty f(k/n)\cdot\frac{(nx)^k}{k!}e^{-nx}\right),$$ as desired!