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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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    alexlo: look over the theorem's hypotheses again. What do these ask you to prove in this case?2012-11-09
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    That there is a function $g(z)$ such that $|f_n(z)|\leq g(z)$ for all $n$ and $\sum g(z) < \infty$, where $f_n(z)=(\dots)$. Unfortunately I haven't found a suitable $g$ yet... My problem is to make it independent of $n$.2012-11-09
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    How do you show the displayed result after "I have already established"? I asked this because maybe it may help you to find a bound independent on $n$ of the absolute value of each term.2012-11-09
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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

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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}