I have trouble applying the Dominated Convergence Theorem in the following situation:
The task is to show that, for $z\in\mathbb{R}$ $$\sum_{k=0}^n (-1)^k \binom{n}{k} \left(1-\frac{k}{n}\right)^{\lfloor n(\log{n}+z) \rfloor} \to e^{-e^{-z}} \text{ as } n \to \infty$$
I have already established that, for a fixed $k\geq 0$ $$(-1)^k \binom{n}{k} \left(1-\frac{k}{n}\right)^{\lfloor n(\log{n}+z) \rfloor} \to \frac{(-e^{-z})^k}{k!} \text{ as } n \to \infty$$
but now I don't see how to justify the limit exchange using dominated convergence to conclude
$$ \lim_{n\to \infty} \sum_{k=0}^n (\dots) = \sum_{k=0}^\infty \lim_{n\to \infty} (\dots) = \sum_{k=0}^\infty \frac{(-e^{-z})^k}{k!} = e^{-e^{-z}}$$