In the post Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$, it is shown that $$\lim_{n\to\infty}e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}=\frac{1}{2}.$$ This limit evaluates differently from $$\lim_{n\to\infty}e^{-n}\sum_{k=0}^{\infty}\frac{n^k}{k!}=\lim_{n\to\infty}e^{-n}\cdot e^n=1,$$ and this suggests that the statement $$\lim_{n\to\infty}\sum_{k=0}^{n}f(n,k)=\lim_{n\to\infty}\sum_{k=0}^{\infty}f(n,k)$$ does not hold in general. However, I would like to know whether there exists some criterion on $f(n,k)$ that guarantees this statement is true, and if there could exist some function $f(n,k)$ such that both limits are finite but not equal.
Edit: As suggested by @Gary, the function $f(n,k)=\frac{e^{-n}\cdot n^k}{k!}$ (as shown in the linked post works). Is it possible to tell when this is true in general, without evaluating the limits explicitly, and does there exist any relation between the two limits?
