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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}}$

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    $(-1)^k \binom{n}{k} \left(1-\frac{k}{n}\right)^{\lfloor n(\log{n}+z) \rfloor} = (-1)^k \frac{n(n-1)\cdots (n-k+1)}{k!} \exp{\{\lfloor n(\log{n}+z) \rfloor (-\frac{k}{n}+o(n^{-2}))\}} \to (-1)^k \frac{n^k}{k!} \exp{\{-k\log{n} -kz\}} = \frac{(-e^{-z})^k}{k!}$ I still don't see it. It seems I'm completely blind to this...2012-11-09

1 Answers 1

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Let $a_{n,k}:=(-1)^k \binom{n}{k} \left(1-\frac{k}{n}\right)^{\lfloor n(\log{n}+z) \rfloor}.$ We have to show that for each $k$, we can find a constant $c_k$ such that for all $n$, $|a_{n,k}|\leqslant c_k\mbox{ and }\sum_{k=1}^{+\infty}c_k<\infty.$ As $\lfloor n(\log{n}+z) \rfloor\geqslant n(\log n+z)-1$ and $1-\frac kn\leqslant 1$, we have for $n>k$ that \begin{align}|a_{n,k}|&\leq \frac{n^k}{k!}\frac 1{1-\frac kn}\exp\left((n\log n+nx)\log\left(1-\frac kn\right)\right)\\ &\leq \frac{n^k}{k!}(k+1)\exp\left(-\frac kn(n\log n+nx)\right)\\ &=(k+1) \frac{n^k}{k!}\exp(-k\log n-kx)\\ &=\frac{k+1}{k!}e^{-kx}. \end{align}