Let $f(x)$ be a real-valued twice continuously differentiable function, and considered the below double sum $$F(t,f(x)):=\dfrac{1}{t}\Big(\sum\_{k=0}^{\infty}\sum\_{m=0}^{\infty}f(x+(k-m)/\sqrt{n})\dfrac{t^{k+m}}{2^{k+m}k!m!}e^{-t}-f(x)\Big).$$ I want to compute the limit of $F(t,f(x))$ when $t\rightarrow 0$, but I don't even know how to start....


My idea was to write this double sum into two parts, the first part may be our desired answer and the second part may be zero when $t\rightarrow 0$, but I don't know what to do to analyze such a complicated sum...


Any idea?



Analyze a complicated double summation
$$F(t,f(x))=\dfrac{1}{t}\Big(\sum\_{k=0}^{\infty}\sum\_{m=0}^{\infty}f(x+(k-m)/\sqrt{n})\dfrac{t^{k+m}}{2^{k+m}k!m!}e^{-t}-f(x)\Big)$$
only the terms $k=0,m=0$, $k=0,m=1$, $k=1,m=0$ survive in the limit $t\rightarrow 0$, the other terms vanishing at least linearly in $t$,
$$\lim\_{t\rightarrow 0}F(t,f(x))=-f(x)+\tfrac{1}{2}f(x+1/\sqrt n)+\tfrac{1}{2}f(x-1/\sqrt n),$$
where I used that
$$\lim\_{t\rightarrow 0}\frac{e^{-t}-1}{t}=-1.$$


