Ooh, I know this one. Just convert it to definite integral!

This proof needs some extra justification. It is true that $\big(1 - \tfrac{k}{n}\big)^n \to e^{-k}$ as $n \to \infty$ for any $k \ge 0$ , but to deduce from this that $\lim_{n \to \infty} \sum_{k=0}^{n-1} \big(1 - \tfrac{k}{n}\big)^n \; = \; \sum_{k=0}^\infty e^{-k}$ and hence obtain the desired result, is nontrivial - the individual summands converge, but we are combining more and more of them at a time, and this could makes things go wrong.

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As an example of how things can go wrong, it is certainly true that $\tfrac{k}{n} \to 0$ as $n \to \infty$ for any $k$ . But we cannot deduce from this that $\sum_{k=1}^n \tfrac{k}{n} \; \to \; \sum_{k=1}^\infty 0 \; = \; 0 \hspace{2cm} n \to \infty$ since, of course, $\sum_{k=1}^n \tfrac{k}{n} \; = \; \tfrac12(n+1) \to \infty \hspace{2cm} n \to \infty$

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We need a discrete version of the Dominated Convergence Theorem. In this case the observation that $\big(1 - \tfrac{k}{n}\big)^n \; \le \; e^{-k} \hspace{2cm} 0 \le k \le n-1$ is what is needed.

Given $\varepsilon > 0$ , find $K$ such that $\sum_{k=K}^\infty e^{-k} < \tfrac12\varepsilon$ . Thus $0 \; < \; \frac{e}{e-1} - \sum_{k=0}^{n-1} \big(1 - \tfrac{k}{n}\big)^n \; \le \; \sum_{k=0}^{K-1} \left(e^{-k} - \big(1 - \tfrac{k}{n}\big)^n\right) + \sum_{k=K}^{n-1} e^{-k} \; < \; \sum_{k=0}^{K-1} \left(e^{-k} - \big(1 - \tfrac{k}{n}\big)^n\right) + \tfrac12\varepsilon$ for all $n > K$ . We can now find $N_k$ such that $0 \; < \; e^{-k} - \big(1 - \tfrac{k}{n}\big)^n < \tfrac{1}{2K}\varepsilon \hspace{2cm} n > N_k$ for all $0 \le k \le K-1$ , and hence we deduce that $0 \; < \; \frac{e}{e-1} - \sum_{k=0}^{n-1}\big(1 - \tfrac{k}{n}\big)^n \; < \; \varepsilon \hspace{2cm} n > \mathrm{max}(K,N_0,N_1,...,N_{K-1})$ This proves the result.

We need the fact that the terms $\big(1 - \tfrac{k}{n}\big)^n$ are dominated uniformly in $n$ by $e^{-k}$ to be able to make this argument.